THE  LIBRARY 

OF 

THE  UNIVERSITY 
OF  CALIFORNIA 

LOS  ANGELES 


GIFT  OF 

John  S. 


KEY  AND   SUPPLEMENT 


TO 


ELEMENTARY  MECHANICS. 


BY 


DE  VOLSON    WOOD, 

PROCESSOR  MATHEMATICS  AND  MECHANICS   IN   BTETEN8  INSTITUTE 
OF  TECHNOLOGY. 


FIRST    EDITION, 

JOHN  S.  PRELL 

Gtiil  &  Mechanical  Engineer* 

SAN  FRAJS  CISCO,  CAL, 

FIRST     THOUSAND. 


NEW   YORK: 

JOHN   WILEY  &   SONS,   PUBLISHERS, 

53   EAST   TENTH   STREET. 

1894. 


COPYRIGHT,  1882, 
BY  DE  VOLSON  WOOD. 


MH»  or  j.  j.  LITTII  4  co.( 
it  10  M.ASIO*  rLACE,  NE«  rear. 


Engineering 

Library 


NOTICE. 


THIS  work  contains  not  only  solutions  of  the  exam* 
pies  and  answers  to  the  Exercises  of  the  Elementary 
Mechanics  of  the  author ;  but  also  additions  to  the  text, 
and  other  new  matter  intended  to  interest  the  general 
student  of  mechanics,  and  be  of  service  to  teachers  of 
the  science  in  connection  with  any  other  text  book. 
Valuable  assistance  has  been  rendered  in  correcting 
proofs  by  Professor  H.  A.  Howe,  of  the  University  of 
Denver. 

DEYOLSON  WOOD. 


737390 

Engineering 
Library 


KEY  AND  SUPPLEMENT 


TO 


ELEMENTARY  MECHANICS. 


PAGE  1,  ARTICLE  1. — Motion  is  a  change  of  position. 
Motion  is  determined  by  the  relative  position  of  bodies 
at  different  times.  If  bodies  retain  the  same  relative 
position  during  successive  times,  they  are  said  to  have 
no  motion  in  reference  to  each  other;  in  other  words, 
they  are  said  to  be  at  rest  in  reference  to  each  other. 
All  bodies  of  which  we  have  any  knowledge  are  in 
motion ;  hence  all  motion  is,  so  far  as  we  know,  rel- 
ative. Absolute  motion  implies  reference  to  a  point 
absolutely  at  rest,  but  as  no  such  point  is  known,  such 
motion  has  only  an  ideal  existence. 

PAGE  2,  ART.  6. — No  definition  of  space  will  give  a  bet- 
ter idea  than  that  obtained  by  experience.  Metaphysi- 
cians have  indulged  in  speculations  in  regard  to  its  na- 
ture, but  they  are  able  to  assert  with  certainty  only 
that  it  has  the  property  of  extension.  Descartes  taught 
that  the  properties  of  extension,  known  as  length, 
breadth,  and  thickness,  were  solely  properties  of  mat- 
ter, and  hence  when  a  body  was  removed  no  space 
remained  in  the  place  formerly  occupied  by  it.  So 
far  as  we  know,  no  space  exists  which  is  perfectly  de- 
ll 


KEY  AND  SUPPLEMENT 

void  of  matter.  Perfectly  void  space  is  an  ideality  ; 
still  modern  philosophy  distinguishes  between  the 
thing  contained,  and  that  which  contains  it — between 
matter  and  the  place  occupied  by  matter.  It  abstracts 
(so  to  speak)  space  from  matter,  and,  in  a  measure, 
matter  from  space.  It  seems  impossible  to  conceive 
of  matter  not  occupying  space,  but  it  is  not  difficult  to 
conceive  of  a  given  quantity  of  matter  as  occupying 
a  very  small  or  a  very  large  space.  We  are  able  to  con- 
sider matter  in  the  abstract  without  considering  the 
dimensions  of  the  space  it  occupies  ;  and  also  we  may 
consider  space  in  the  abstract  as  not  including  matter. 
The  latter  is  called  absolute  space;  it  is  conceived 
as  remaining  always  similar  to  itself  and  immovable. 

Time  is  duration.  We  gain  a  knowledge  of  it  by 
the  order  of  events.  Every  event  has  its  place  in 
time  and  space,  and  by  means  of  memory  we  gain  a 
knowledge  of  the  order  in  which  events  occur.  "With- 
out memory  we  would  gain  %  experience  no  knowl- 
edge of  time.  Sir  Isaac  Newton  considered  mathe- 
matical time  as  flowing  at  a  uniform  rate,  unaffected 
by  the  motions  of  material  things.  This  idea  induced 
him  to  call  his  new  ealcnlns  fluxions. 

Rate  refers  to  some  unit  as  a  standard.  Thus,  to 
illustrate,  rate  of  interest  is  a  certain  amount  of  money 
paid  for  the  use  of  one  dollar ;  passenger  rates  refer 
to  the  amount  paid  for  one  passenger  ;  rate  of  shipping 
per  ton  is  the  amount  paid  for  carrying  one  ton  ;  rate 
of  motion  is  the  space  passed  over  in  one  second,  one 
minute,  one  hour,  one  day,  or  one  of  any  other  unit. 
The  term  velocity  is  simply  the  equivalent  of  the  rate 
of  motion.  Angular  velocity  is  rate  of  angular  motion 


TO  ELEMENTARY  MECHANICS.  3 

(p.  4,  Art.  12).  Acceleration  is  the  rate  of  change  of 
velocity,  being  the  amount  of  change  in  the  velocity 
for  one  second,  or  one  of  any  other  unit  (p.  10,  Art. 
22).  Mechanical  power  is  the  rate  of  doing  work  (p. 
55,  Art.  99).  Rates  may  involve  two  units.  Thus, 
rate  per  ton- mile  implies  a  certain  amount  paid  for 
one  ton  for  one  mile ;  passenger  rates  are  often  an 
amount  for  one  person  for  one  mile  ;  mechanical  power, 
or  rate  of  doing  work,  is  the  amount  of  work  done  in 
one  foot  for  one  second,  or  by  one  pound  for  one  sec- 
ond, etc.  Hate  is  a  thing  used  for  measuring  quantities, 
as  a  yard-stick  is  used  for  measuring  cloth,  a  chain 
for  measuring  land,  the  pound  for  weighing  groceries, 
ton  for  measuring  merchandise,  etc. 

PAGE  4,  ART.  12. — The  definition  here  given  for  rotary 
motion  is  applicable  to  the 
case  where  the  motion  is  in 
a  curved  path  not  circular, 
as  CD.  But  the  analysis 
given  in  the  text  is  not  applicable  to  this  case. 

PAGE  6,  ART.  14. — Just  after  Fig.  5,  for  If  two  ve- 
locities, etc.,  read  If  two  concurrent  simultaneous 
velocities,  etc. 

PAGE  8,  ART.  20. — Speaking  of  the  rotation  of  the 
moon,  suggests  an  interesting  question  in  practical 
mechanics.  If  the  wheel 
£  rolls  around  on  the  cir- 
cumference of  an  equal 
wheel  A,  will  the  former 
turn  once  or  twice  on  its 
own  axis? 

Mark  the  point  a  which,  initially,  is  in  contact  with 


KEY  AND  SUPPLEMENT 


the  wheel  A,  and  roll  B  half  around  A  ;  it  will  be 
observed  that  the  marked  point  will  then  be  at  the 
left  of  the  centre  of  B,  as  it  was  at  the  start.  Con- 
tinuing the  rolling,  the  point  a  will  again  be  at  the 
left  of  the  centre  when  B  has  gone  completely  around 
A  ;  hence  it  is  sometimes  asserted  that  the  wheel  B 
has  turned  twice  on  its  axis.  But  we  wish  to  show 
that  it  has  turned  but  once  on  its  own  axis,  and  the 
whole  wheel  has  been  rotated  once  about  the  axis  of 
the  wheel  A.  Let  the  axis  of  the  wheels  be  at  right 

angles  with  each  other,  then 
it  will  be  evident,  from 
mere  inspection,  that  when 
B  has  turned  once  on  its 
axis  it  will  have  gone  once 
around  A.  Thus  B  will  have 
gone  once  around  the  axis  of 
A,  and  once  about  its  own 
axis.  Next,  incline  the  axis 
of  B  upward,  so  as  to  ap- 
proach a  parallelism  to  A, 
and  the  same  result  will  be 
seen  from  mere  inspection, 
and  it  will  continue  to  remain  evident  as  it  be- 
comes nearer  and  nearer  parallel,  and  when  they 
become  actually  parallel,  the  same  condition  will 
hold  true.  Hence,  in  the  former  figure,  the  wheel 
B  will  turn  but  once  on  its  own  axis  in  rolling  once 
around  A.  The  same  result  may  be  shown  in  an- 
other way.  Let  a  block  be  placed  at  a  facing  a  mark 
on  the  axis  of  B,  and  conceive  this  axis  to  be  rigidly 
connected  with  the  axis  of  A  while  the  wheel  B  is 


TO  ELEMENTARY  MECHANICS.  5 

free  to  turn  on  its  own  axis.  In  this  way  the  axis  of 
B  will  be  carried  bodily  about  A.  When  B  has  rolled 
half  around  A,  it  will  be  found  that  the  block  will 
face  the  same  direction  in  space — say  towards  the 
east — but  that  it  will  not  face  the  mark  on  the  axis, 
for  the  mark  will  be  on  the  opposite  side  of  the  axis. 
Continuing  the  rotation,  it  will  be  found  that  the 
block  will  face  the  mark  only  once  at  each  revolution 
about  A. 

Similarly,  the  moon  turns  but  once  on  its  own  axis 
in  one  revolution  about  the  earth,  but  the  rotation 
about  the  two  centres  are  not  exactly  coincident ;  for 
it  is  found  by  observation,  that  in  some  parts  of  the 
orbit  more  of  the  surface  of  the  moon  is  seen  on  the 
eastern  (or  western)  side  than  in  other  parts  of  the 
orbit ;  thus  showing  that  the  rotation  about  the  earth 
is  sometimes  faster,  and  at  other  times  slower  than  the 
rotation  of  the  moon  about  its  own  axis.  This  phe- 
nomenon is  called  Libration. 

EXERCISES. 
PAGE  9. 

1.  4^-  miles. 

2.  The  former. 

3.  66  feet  per  second. 

4.  17  feet ;  Vln  feet. 

40  x  5280 
5"  4°°  *     60x60     =  ^  SCCOnds' 

„    200  x  2zr                                200  x  360          10_A 
—^ =  6|7r  m  arc  ;  or ^— 

degrees. 
7.  V&  +  2"  =  Vl3~=  3-605  +  miles  per  hour. 


6  KEF  AND  SUPPLEMENT 

PAGE  10. 


8.   1/15'  4-  =  ^3961  =  20-98  ft.  per  sec- 


ond. 
9.  0-0009J. 

10.  See  Article  14.  v  =  A/5'  +  10"  +  100  cos  60° 
=  Vl25  +  50  =  V175  =  13-22  +  feet  ;  hence 
the  distance  between  them  in  two  seconds  will 
be  2  x  13-22  +  =  26-44  +  feet. 

PAGE  10,  ART.  22.  —  Observe  that  acceleration  is  not  the 
rate  of  change  of  motion,  but  the  rate  of  change  of 
the  rate  of  motion.  It  is  the  rate  of  change  of  a  rate. 
Tlifi  rate  of  change  is  usually  measured  in  the  same 
units  as  the  rate  of  motion.  If  one  is  in  feet  per  sec- 
ond, the  other  is  also.  It  is  possible  to  conceive  of 
mixed  units.  Thus  in  the  case  of  falling  bodies,  the 
velocity  at  the  end  of  the  first  second  is  16-^  feet,  and 
the  acceleration  is  643|  yards  per  minute. 

Strive  to  get  a  clear  conception  of  the  meaning  of 
acceleration  and  of  its  measure.     It  is  one  of  the  ele- 
ments of  the  absolute  measure  of  force. 
PAGE  13,  ART.  26.  —  The  expression  "  The  locus  of  these 
,\  points  will  be  a  parabola,"  means 

that  if  any  number  of  points  in 
the  path  be  determined  in  the 
same  manner,  they  will  all  be 
in  the  arc  of  a  parabola. 

A  parabola   is  a  curve  which 
may  be  cut    from  a  right  cone 
by  a  plane  parallel  to  one  of  its 
elements.     (See  Author's  Coordinate  Geometry.) 


TO  ELEMENTARY  MECHANICS.  7 

EXAMPLES. 

1.  Space  250  feet  ;  velocity  50  feet  per  second. 
PAGE  14. 

2.  /  =  25  feet  ;  *  =  12£  feet. 

3.  During  4  seconds  the  space  will  be  s  =  %  of  32 

x  4"  (Art.  14),  and  during  3  seconds  the  space 
will  be  \  of  32  x  3*  ;  hence  during  the  4th  second 
the  space  will  be  J  x  32  (4a  -  3s)  =  112  feet. 

4.  By  means  of  equation  (1),  Art.  25,  find  6£  feet 

per  second. 


5.  t=  Vl2^5  =  3-533  +  seconds. 

6.  /=  (20  -T-  120)  3-28  =  0-54f  feet  per  second. 

7.  32|  -5-  3-28  =  9-8  metres  per  second. 

PAGE  15.  —  Matter  and  force  are  two  grand  realities  of 
the  external  world,  and  of  these  we  know  nothing 
directly.  Our  knowledge  of  the  former  is  confined 
to  its  properties,  and  of  the  other  to  its  laws  of  action. 
But  we  have  no  reason  to  believe  that  one  exists  in- 
dependently of  the  other.  In  our  earlier  experiences 
matter  is  conceived  to  be  hard,  gross,  and  unyielding  ; 
but  later  we  find  that  it  is  yielding,  and  that  many 
solids,  as  iron  and  lead,  may  be  changed  to  liquids  by 
heat,  and  that  liquids  may  be  changed  to  gases  —  so 
that  matter  is  proved  to  be  more  or  less  viscid  or  at- 
tenuated. Solids  are  porous.  Changes  of  form  are 
effected  by  forces,  so  that  some  metaphysicians  have 
reasoned  that  there  may  possibly  be  no  gross  matter, 
but,  instead  thereof,  those  things  which  we  consider 
as  bodies  are  only  aggregations  of  forces.  On  the 
other  hand,  all  investigations  in  mechanics  proceed 
on  the  hypothesis  that  matter  is  in  no  sense  a  force, 


8  KEY  AND   SUPPLEMENT 

or  an  aggregation  of  forces,  but  that  it  is  something 
distinct  from  force,  something  npori  which  force  acts. 
In  the  case  of  attraction,  the  matter  in  two  bodies 
may  remain  constant,  while  the  force  exerted  by  each 
upon  the  other  will  depend  upon  the  distance  be- 
tween them.  It  is  true,  however,  that  we  gain  a 
knowledge  of  matter  only  through  the  action  of  forces. 
Every  avenue  to  the  mind  through  the  senses  is  an 
agent  for  transmitting  the  result  of  the  action  of  cer- 
tain forces,  and  the  very  act  of  transmission  brings 
into  play  certain  forces. 

PAGE  16,  ART.  28. — Mathematics  applied  to  the  laws  of 
physical  science  enables  us  to  determine  magnitudes 
which  far  transcend  the  powers  of  accurate  measure- 
ment or  even  of  conception. 

Sir  Win.  Thompson  gives  four  methods  for  ascer- 
taining the  mean  distance  between  molecules. 

Optical  dynamics. 

Contact  electricity  of  metals. 

Capillary  attraction. 

Kinetic  theory  of  gases. 

"  Optical  dynamics  leaves  no  alternative  but  to  admit 
that  the  diameter  of  a  molecule,  or  the  distance  from 
the  centre  of  a  molecule  to  the  centre  of  a  contiguous 
molecule  in  glass,  water,  or  any  other  of  our  transpa- 
rent liquids  and  solids,  exceeds  one  ten-thousandth  of 
the  wave-length  of  light,  or  a  two-hundred-millionth 
of  a  centimetre "  (sooooVooooo  of  a  metre). 

"  However  difficult  it  may  be  even  to  imagine  what 
kind  of  thing  the  molecule  is,  we  may  regard  it  as  an 
established  truth  of  science  that  a  gas  consists  of  mov- 
ing molecules  disturbed  from  rectilineal  paths  and 


TO  ELEMENTARY  MECHANICS.  9 

constant  velocities  by  collisions  or  mutual  influences, 
so  rare  that  the  mean  length  of  proximately  rectilineal 
portions  of  the  path  of  each  molecule  is  many  times 
greater  than  the  average  distance  from  the  centre  of 
each  molecule  to  the  centre  of  the  molecule  nearest 
it  at  any  time.  If  for  a  moment  we  suppose  the  mole- 
cules to  be  hard  elastic  globes  all  of  one  size,  influenc- 
ing one  another  only  through  actual  contact,  we  have 
for  each  molecule  simply  a  zigzag  path  composed  of 
rectilineal  portions,  with  abrupt  changes  of  direction." 
"  If  the  particles  were  hard  elastic  globes,  the  aver- 
age time  from  collision  to  collision  would  be  inversely 
as  the  average  velocity  of  the  particle.  But  Max- 
well's experiments  on  the  variation  of  the  viscosities 
of  gases  with  change  of  temperature  prove  that  the 
mean  time  from  collision  to  collision  is  independent 
of  the  velocity,  if  we  give  the  name  collision  to  those 
mutual  actions  only  which  produce  something  more 
than  a  certain  specified  degree  of  deflection  of  the 
line  of  motion.  This  law  could  be  fulfilled  by  soft 
elastic  particles  (globular  or  not  globular),  but  not  by 
hard  elastic  globes."  "  By  Joule,  Maxwell,  and  Clau- 
sius  we  know  that  the  average  velocity  of  the  mole- 
cules of  oxygen,  or  nitrogen,  or  common  air,  at  or- 
dinary atmospheric  temperature  and  pressure,  is  about 
50,000  centimetres  per  second  (500  metres  per  sec- 
ond, or  about  1,600  feet  per  second),  and  the  average 
time  from  collision  to  collision  a  five-thousand- 
millionth  of  a  second  (goooo^oooo)-  Hence  the  aver- 
age length  of  path  of  each  molecule  between  collis- 
ions is  about  Yo-oVoir  °f  a  centimetre  "  (iTnnnnnnr  °^  a 
metre). 


10  KEY  AND  SUPPLEMENT 

"  The  experiments  of  Cagniard  de  la  Tour,  Faraday, 
Regnault,  and  Andrews  on  the  condensation  of  gases 
do  not  allow  us  to  believe  that  any  of  the  ordinary 
gases  could  be  made  forty  thousand  times  denser  than 
at  ordinary  atmospheric  pressure  and  temperature 
without  reducing  the  whole  volume  to  something  less 
than  the  sum  of  the  volumes  of  the  gaseous  molecules, 
as  now  defined. 

"  Hence,  according  to  Clausius,  the  average  length 
of  path  from  collision  to  collision  cannot  be  more 
than  five  thousand  times  the  diameter  of  the  gaseous 
molecule  ;  and  the  number  of  molecules  in  unit  of 
volume  cannot  exceed  25,000,000  divided  by  the  vol- 
ume of  a  globe  whose  radius  is  that  average  length  of 


path.  Taking  now  the  estimated  y^Vur  of  a  centime- 
tre for  the  average  length  of  path  from  collision  to 
collision  wo  conclude  that  the  diameter  of  the  gaseous 
molecules  cannot  be  less  than  sooooVopir  °f  a  centime- 
tre  Cnnnnf  crznnnnr  °f  a  metre)  ;  nor  the  number  of 
molecules  in  a  cubic  centimetre  of  the  gas  (at  ordi- 
nary density)  greater  than  6,000,000,000,000,000,000,- 
000." 

"The  densities  of  known  liquids  and  solids  are 
from  five  hundred  to  sixteen  thousand  times  that 
of  atmospheric  air  at  ordinary  pressure  and  tempera- 
ture :  and,  therefore,  the  number  of  molecules  in  a 
cubic  centimetre  may  be  from  3  x  10**  to  10s  6  (that  is, 
from  three  million  million  million  million,  to  a 
hundred  million  million  million  million).  From 
this  the  distance  from  center  to  nearest  center  in 
solids  and  liquids  may  be  estimated  at  from  ^OTJOTTFITU 
to  iinnnhnnnr  of  a  centimetre  (rnn.uVu.nnn;  to 


TO  ELEMENTARY  MECHANICS.  H 

46060000UUU  of  a  metre).  The  four  lines  of  argu- 
ment lead  all  to  substantially  the  same  estimate  of 
the  dimensions  of  molecular  structure.  Jointly  they 
establish  with  what  we  cannot  but  regard  as  a  very 
high  degree  of  probability  the  conclusion  that,  in 
any  ordinary  liquid,  transparent,  solid,  or  seemingly 
opaque  solid,  the  mean  distance  between  the  centers 
of  contiguous  molecules  is  less  than  the  hundred- 
millionth,  and  greater  than  the  two  thousand-millionth 
of  a  centimetre.  To  form  some  conception  of  the 
coarse-grainedness  indicated  by  this  conclusion,  im- 
agine a  rain  drop,  or  a  globe  of  glass  as  large  as  a  pea, 
to  be  magnified  up  to  the  size  of  the  earth,  each  con- 
stituent molecule  being  magnified  in  the  same  pro- 
portion. The  magnified  structure  would  be  coarser- 
grained  than  a  heap  of  small  shot,  but  probably  less 
coarse-grained  than  a  heap  of  cricket-balls."  (Ex- 
tracts from  a  paper  by  Prof.  Sir  "Wm.  Thomson  on 
the  size  of  Atoms,  Am.  Jour,  of  Science  and  Art. 
18TO,  vol.  ii.,  pp.  38-45.) 

Mr.  1ST.  D.  C.  Hodges,  in  an  article  on  the  size  of 
molecules  (Phil.  Mag.  and  Jour,  of  Science,  1S79, 
vol.  ii.,  p.  74),  says :  "  If  we  consider  unit  mass  of 
water,  the  expenditure  on  it  of  an  amount  of  energy 
equivalent  to  636.7  units  of  heat  will  convert  it  from 
water  at  zero  into  steam  at  100°.  I  am  going  to  con- 
sider this  conversion  into  steam  as  a  breaking-up  of 
the  water  into  a  large  number  of  small  parts,  the  total 
surface  of  which  will  be  much  greater  than  that  of 
the  water  originally.  To  increase  the  surface  of  a 
quantity  of  water  by  one  square  centimetre  requires 
the  use  of  .000825  metre  gramme  of  work.  The  total 


12  KEY  AND  SUPPLEMENT 

superficial  area  of  all  the  parts,  supposing  them  spher- 
ical, "will  be  4  7t  r*N)  the  number  of  parts  being  W. 
The  work  done  in  dividing  the  water  will  be  4  it  r* 
N  x  .000825.  For  the  volume  of  all  the  parts  we 
have  f  n  r*N.  This  volume  is,  in  accordance  with 
the  requirements  of  the  kinetic  theory  of  gases,  about 
3,000  of  the  total  volume  of  the  water.  The  volume 
of  the  steam  is  1,752  times  the  original  unit  volume 
of  water.  Hence — 

i  nr*N  3000  =  1752 

4  n  r*  ^.000825  =  G36.7423. 

One  unit  of  heat  equals  423  units  of  work  (in  French 
units);  solving  these  equations  for  '/•  and  JV,  we  get 
r  =  .000000005  centimetre  (or  diameter  =  .00000001 
centimetre  =  nnnnrWinnnF  metre),  a  quantity  closely 
corresponding  with  the  previous  results  of  Sir  "Wm. 
Thomson,  Maxwell,  and  others  ;  and  N  equals  9000 
(million)8,  or  for  the  number  in  one  cubic  centime- 
tre 5  to  6  (million)3." 

The  extreme  tenuity  of  a  gas  is  further  shown  by 
the  following  extract  taken  from  the  Beiblatter  zu 
den  Annalen  der  Phymk  und  Chemie,  1879,  No.  2, 
p.  59.  "At  0°C.  and  IQOmni  pressure  a  cubic  cen- 
timetre (.061  cubic  inch)  holds  nearly  one  hundred 
trillions  of  gas  molecules.  Under  these  conditions 
the  molecules  themselves  fill  nearly  the  -j-^  of  the 
space  occupied  by  the  gas.  The  absolute  weight  of  a 

15 

hydrogen  molecule  is  represented  by  — -^  <?,  (g  in  me- 
tres)." 
Mr.  G.  J.  Stoney,  in  an  article  on  Polarization  on 


TO  ELEMENTS Rl   MECHANICS.  13 

Stress  in  Gases  (Phil.  Mag.  and  Jour,  of  Science,  1878, 
vol.  ii.,  p.  407),  says  that  "  the  number  of  molecules  in 
a  cubic  millimetre  of  atmospheric  air  is  about  1018 
(  -  1,000,000,000,000,000,000),*  and  that  the  average 
distance  between  them  is  about  TOTFTO'Q  o  o  u  o  °f  a  metre. 
The  average  striking  distance  (i.  e.  the  average  length 
of  path  between  encounters)  of  the  molecules  is  about 
TS 00*0000  °f  a  metre.  The  average  velocity  at  ordi- 
nary temperature  may  be  taken  as  500  metres  per  sec- 
ond (1,600  feet  per  second),  and  the  molecules  meet 
with  so  many  encounters,  that  the  direction  of  the 
path  of  each  is  changed  somewhere  about  10,000,000,- 
000  times  every  second."  In  one  movement  the  par- 
ticle travels  ^oiroWcnr  °^  a  metre>  or  soootr  °f  a  milli- 
metre ;  t  and  it  makes  this  movement  in  10oo61u-u6o66 
of  a  second.  Now  the  wave  length  of  a  chemical 
ray  is  about  ^-^  °f  a  millimetre,  hence  we  find  that 
the  molecule  of  air  travels  through  a  distance  which 
is  one-fourth  as  long  as  the  length  of  this  particular 
wave  in  this  fraction  of  a  second.  J 

According  to  Pouillet,  the  mechanical  energy  of  a 
cubic  mile  of  sun  light  at  the  earth  equals  12,050  ft.  Ibs. 

*  Clausius's  limit  is  6,000  times  this  amount. 

f  Clausius's  estimate  is  one-half  this  value. 

\  Mr.  E.  H.  Cook,  in  an  article  on  the  Existence  of  the  Lumi- 
niferous  Ether  (Phil.  Mag.  and  Jour,  of  Science,  1879,  vol.  I,  p. 
235),  after  quoting  the  above  figures  from  Mr.  Stoney's  paper, 
adds  :  "Then  in  one  moment  the  particle  travels  roTJooTHT  °f  a 
metre  in  TuoHootfiToocT  of  a  second;"  also  in  his  deductions  he 
adds  :  "  hence  we  find  that  the  molecule  of  air  travels  through  a 
distance  which  is  more  than  twice  as  long  as  the  length  of  this 
particular  wave  in  this  fraction  of  a  second."  The  reader  can 
easily  see  that  this  deduction  is  erroneous. 


KEY  AND  SUPPLEMENT 


(Phil.  Mag.,  1855,  vol.  ix.,  p.  39)  ;  accepting  which 
Sir  Wm.  Thomson  calculated  the  weight  in  pounds 
of  a  cubic  foot  of  the  ether  of  space  (that  which  is 
instrumental  in  transmitting  light,  and  called  ether) 
by  the  formula  : 


where  g  =  32£  (acceleration  per  second  of  gravity), 
F=the  velocity  of  light  per  second,  being  about 
192,000  miles  per  second,  and  n  =  -g^,  being  the  ratio 
of  the  greatest  velocity  of  a  rotating  particle  to  the 
velocity  of  light.  Herr1  Glan  asserts  that  n  is  not 
constant,  that  he  found  n  =  ^V  in  one  case,  and  y^Vs" 
in  another.  (Am.  Jour,  of  Arts  and  Sciences,  1879, 
vol.  xviii.,  404.  Annalen  der  Physilt  und  Chemie, 
No.  8,  1879,  p.  584.)  Assuming  n  =  7V>  we  find  that 
a  cubic  foot  would  weigh  about 

w_    83  x  32|  x  5Q2 
~  (52bO  x  192000)8 

1-5   ,, 
=  ioTo  ^.nearly; 

and  for  the  weight  of  a  cubic  mile  -—  -  pound.     The 

weight  of  a  volume  of  the  size  of  the  earth  would  be 
about  240  pounds.  Admitting  this  result,  it  follows 
that  a  sphere  equal  in  diameter  to  the  diameter  of  the 
earth's  orbit  (or  say  190,000,000  miles)  will  contain 
an  amount  of  ethereal  matter  nearly  equal  to  j^Vo- 
of  that  of  the  mass  of  the  earth. 

Probable  tension  of  the  ether  of  space.     The  above 


TO  ELEMENTARY  MECHANICS.  15 

results  combined  with  one  or  two  other  plausible 
assumptions  enable  us  to  find  a  probable  value  of 
the  tension  of  the  light  ether.  As  stated  above, 
the  velocity  of  the  particles  of  air  producing  a 
pressure  of  15  Ibs.  per  square  inch,  is  about  1,600 
feet  per  second,  which  is  about  50  per  cent,  more 
than  the  velocity  of  sound  in  air.  Assuming  now 
that  the  normal  velocity  of  the  ether  particles  is 
somewhat  more  than  the  velocity  of  light — or,  more 
definitely,  that  it  is  195,000  miles  per  second,  and 
that  the  tension  is  directly  as  the  masses,  and  also  as 
the  square  of  the  velocities  of  the  particles ;  also  that 

the  weight  of  a  cubic  mile  of  the  ether  is  -^-  lb.,  as 

found  above,  and  observing  that  100  cubic  inches 
of  ordinary  air  weigh  31  grains,  and  7,000  grains 
make  a  pound,  we  have  for  the  pressure  P  in  pounds 
of  the  ether  upon  a  square  inch 

J_ 

iK       /195000  x  5280\8  109 

P  =  15  x  ( — — x 


1600          J       31        1728 
100      - 


=  0-00000185  =          lb.  per  inch  of  section. 

A  Mr.  Preston,  an  English  writer,  in  his  work  on  the 
Physics  of  Ether,  estimates,  or  rather  assumes,  500 
tons  per  square  inch  as  a  probable  inferior  limit  of 
the  pressure  (p.  18),  and,  with  this  as  a  basis,  he  finds 
the  weight  of  a  cubic  mile  of  ether  to  be  about  220  Ibs. 
(p.  120).  But  as  he  has  used  56-5  grains  for  the 
weight  of  a  cubic  foot  of  air  when  it  is  nearly  ten 


IQ  KEY  AND  SUPPLEMENT 

times  this  amount,  he  should  have  found  for  the 
weight  of  a  cubic  mile  of  ether  nearly  one  ton.  His 
assumption,  however,  in  regard  to  pressure  is  quite 
arbitrary,  and  does  not  seem  to  be  well  founded.  It 
seems  improbable  that  there  should  be  so  much  mass 
in  the  ether  of  space.  Even  240  Ibs.  for  a  volume 
equal  to  that  of  the  earth  seems  a  high  value  when  \\Q 
consider  the  amount  that  must  be  displaced  by  the 
planets  while  moving  about  their  orbits. 

Temperature  of  the  Sun.  •The  following  may  be  interesting, 
although  it  does  not  fall  under  the  article  above  referred  to. 
"  The  effective  temperature  of  the  sun  may  be  defined  as  that 
temperature  which  an  incandescent  body  of  the  same  size  placed 
at  the  same  distance  ought  to  have  in  order  to  produce  the  same 
thermal  effect  if  it  had  the  maximum  emissive  power.  If  we 
consider  the  surrounding  temperature  during  the  observation  to 
have  been  about  240°  we  obtain  .  .  .  for  the  effective  tempera- 
ture in  degrees  centigrade  9,905.4° ...  I  think,  then,  that  I  may 
fairly  conclude  that  the  temperature  of  the  sun  is  not  very  dif- 
ferent from  its  effective  temperature,  and  that  it  is  not  less  than 
10,000°,  nor  much  more  than  20,000'  centigrade."  Phil.  Mag., 
1879,  vol.  ii.,  pp.  548-550.  See  also  Am.  Jour.  Sc.  and  Arts, 
1870,  vol.  ii.,  p.  68. 

Sir  Win.  Thomson  calculates  the   mechanical   energy   of  the 

solar  rays  falling  annually  on  a  square  foot  of  land  in  latitude  50° 

to  equal  530,000,000  foot  pounds,  or  396  H.  P.  per  yard  per  day. 

He  finds  tha*,  the  heat  alone  hourly  given   out  by  each  square 

yard  of  the  solar  surface  is  equivalent  to  63,000  horse  power,  and 

-  would  require  then  the  hourly  combustion  of  13,500  Ibs.  of  coal. 

Appleton's  Cyclopaedia,  1868,  vol.  ix.  p.  23. 

PAGE  16,  AET.  29. — A  better  definition  is — Force  is  an 
action  between  bodies — for  this  form  of  the  statement 
recognizes  the  existence  of  at  least  two  bodies  in 
every  action.  A  force  never  acts  upon  one  body 
without  producing  an  equal  opposite  action  upon  an- 


TO  ELEMENTARY  MECHANICS.  17 

other  body.  We  speak  of  the  action  of  a  force  upon 
a  body  because  in  most  cases  the  second  body  is  so 
large,  relatively,  that  the  force  produces  little  or  no 
perceptible  effect  upon  it.  But  according  to  the  law 
of  Universal  Gravitation  each  particle  in  the  universe 
is  attracted  by  every  other  particle  with  a  force  which 
depends  upon  their  masses  and  the  distances  between 
them  ;  hence,  in  a  highly  refined  sense,  it  may  be  said 
that  every  force  producing  motion  involves  every 
body  in  the  universe.  The  entire  universe  of  matter 
is  bound  together  by  a  something — an  action — which 
we  call  Force.  H/very  phenomenon  which  we  witness 
in  the  physical  world  is  the  result  of  force,  acting 
through  space,  or  during  a  certain  time. 

Force  alone  is  stress.  In  other  words  stress  is 
force  abstracted  from  time  and  space.  The  science 
of  Stress  is  the  science  of  Statics.  Stress  is  always 
measured  in  pounds  or  its  equivalent.  If  a  force  pro- 
duces motion,  that  part  of  the  phenomenon  which  is 
abstracted  from  time  and  space  is  stress,  so  that  the 
attractive  force  between  the  earth  and  moon,  or  be- 
tween the  earth  and  sun,  measured  in  pounds,  is 
stress. 

When  force  is  compounded  with  time  or  space  the 
result  is  work,  or  energy,  or  momentum,  as  will  here- 
after be  shown. 

The  following  are  some  definitions  of  force  as  given 
by  different  authors : 

La  Place  says  :  "  The  nature  of  that  singular  mod- 
ification, by  means  of  which  a  body  is  transported 
from  one  place  to  another,  is  now,  and  always  will 
be,  unknown ;  it  is  denoted  by  the  name  of  Force. 


18  KEY  AND  SUPPLEMENT 

We  can  only  ascertain  its  effects,  and  the  laws  of  its 
action."  (Mecanique  Celeste,  p.  1). 

"  .Force  is  an  action  between  bodies,  causing  or 
tending  to  cause  change  in  their  relative  rest  or  mo- 
tion "  (Rankine's  Applied  Mechanics,  p.  15). 

"  Force  is  that  principle  of  which,  considered  sim- 
ply as  a  mechanical  agent,  we  know  but  little  more 
than  that  when  it  is  imparted,  that  is,  put  into,  a 
body,  it  produces  either  motion  alone  ;  or  strain,  with 
or  without  motion."  "  What  is  called  overcoming 
inertia,  is  simply  putting  in  force"  (Trautwine's 
Engineer's  Pocket  Book,  p.  445  and  p.  447).  We 
consider  this  as  a  misuse  of  terms.  We  cannot  safely 
say  that  force  is  put  into  a  body.  When  force  acts 
upon  a  body  free  to  move,  energy  is  put  into  the 
body,  when  force  and  space  are  involved  in  the  result, 
and  time  is  abstracted  (see  text,  p.  66) ;  or  taoinentum 
when  the  elements  involved  are  force  and  time,  space 
being  abstracted  (see  Chap.  Y ).  Although  it  is  too 
early  in  the  text  for  a  full  discussion,  we  add  a  re- 
mark for  the  benefit  of  those  who  have  some  knowl- 
edge of  the  subject.  When  a  constant  force,  F,  acts 
through  a  space,  s,  it  does  the  work  represented  by 
the  product  of  F  and  s,  or  Fs.  If  the  body  upon 
which  it  acts  is  wholly  free,  the  entire  work  will  be 
stored  in  the  body,  and  is  then  called  energy,  the 
measure  of  which  is  %  Mv*;  hence  Fs  =  £  Mv*.  Elimi- 
nating s  by  means  of  Eq.  (2),  p.  12,  of  the  text,  gives 
Ft  =  Mv,  which  is  the  measure  of  the  effect  of  force 
combined  with  time,  and  the  second  member  is  the 
measure  of  the  momentum  (text  p.  78).  It  would  be 
better  to  say  that  force  and  time,  or  force  and  space, 


TO  ELEMENTARY  MECHANICS.  19 

are  put  into  a  body  than  that  force  alone  is  put  into 
it.  "  Nothing  but  force  can  resist  force."  "Matter, 
in  itself,  cannot  resist  force "  (ib.  p.  445).  By  re- 
sistance is  here  understood  to  be  such  a  condition  of 
things  as  that  motion  will  not  result,  and,  in  this 
sense,  these  statements  are  correct. 

"Force  put  into  a  body  "  is,  properly  speaking, put- 
ting it  under  stress,  and  the  body  is  said  to  be  strained, 
but  no  amount  of  internal  stress  will  produce  motion 
of  the  body. 

"  Whatever  changes  the  state  of  a  body  or  the  ele- 
ments of  a  body,  with  respect  to  rest  and  motion,  is 
called  force"  (Bartlett's  Analyt.  Mech.,  p.  17). 

"  Force  is  defined  as  that  which  changes  or  tends 
to  change  a  body's  state  of  rest,  or  motion,  and  any 
given  force  may  be  measured  bjr  the  acceleration  it 
imparts  to  a  gramme."  (Cumraing's  Theory  of  Elec- 
tricity, p.  5). 

"  Force  is  whatever  changes  or  tends  to  change  the 
motion  of  a  body  by  altering  either  its  dimension  or 
its  magnitude ;  and  a  force  acting  on  a  body  is  meas- 
ured by  the  momentum  it  produces  in  its  own  direc- 
tion in  a  unit  of  time."  (Maxwell's  Theory  of  Heat, 
p.  83). 

It  will  appear  that  the  momentum  produced  in  a 
unit  of  time  is  the  same  as  the  acceleration,  and  hence 
the  two  last  definitions  are  equivalent.  "We,  however, 
deem  it  advisable  to  avoid  the  expression  momentum 
produced  because  it  is  liable  to  be  confounded  with  the 
actual  momentum  of  a  body,  although  it  is  not  in- 


20  KEY  AND  SUPPLEMENT 

tended   to   even  imply   the   latter.     Acceleration  is 
specific  and  correct. 

"  Force  is  matter  in  motion,  nothing  more,  nothing 
less ;  the  abstract  idea  of  force  without  matter  is  a 
nonentity."  (Nystrom  On  tlie  Force  of  Falling  Bodies 
and  Dynamics  of  Matter,  p.  20). 

The  preceding  remarks  will  show  that  this  defini- 
tion contains  a  misapplication  of  terms.  Matter  in 
motion  is  either  Energy  or  Momentum,  according  as 
time  or  space  is  abstracted  in  considering  the  ele- 
ments which  enter  into  the  combination. 

"  Force  is  a  mere  name,  but  the  product  of  a  force 
into  the  displacement  of  its  point  of  application  lias 
an  objective  existence." 

"  Force  is  the  rate  at  which  an  agent  does  work 
per  unit  of  length." 

"  The  mere  rate  of  transference  of  energy  per  unit 
of  length  of  that  motion  is,  in  the  present  state  of 
science,  very  conveniently  called  force."  (Lecture  by 
Prof.  G.  P.  Tait,  Nature,  1876,  vol.  xiv.  p.  462). 

In  regard  to  these  views,  we  observe  that  the  name 
applied  to  anything  is  a  mere  name.  In  a  certain 
sense  it  is  an  ideality,  but  generally  the  name  stands 
for  a  reality.  In  this  case,  if  force  is  a  mere  name, 
what  is  the  sense  of  the  remainder  of  the  sentence  ? 
How  can  a  force  have  a  poi,nt  of  application  if  the 
force  is  a  mere  name?  And  granting  that  it  may 
have,  how  can  the  product  have  an  objective  exist- 
ence? 

In  regard  to  the  second  definition,  it  is  analytically 
correct,  but  we  consider  it  rather  as  a  deduction  than 


TO  ELEMENTARY  MECHANICS.  21 

as  a  fundamental  definition.     It  is  shown  on  pages 
52  and  67  that 


or,  in  terras  of  the  calculus, 

Fds  =  d'(lMv*)  =  $Md(tf)  =  dK. 
From  the  former  we  have 


and  from  the  latter 


~~ds  > 

hence,  generally,  force  is  the  rate  of  doing  work  per 
unit  of  length.  But  is  force  merely  a  rate  ?  What 
shall  be  said  of  force  as  a  stress,  where  no  transfer- 
ence of  energy  takes  place  ?  That  this  definition  is 
not  elementary,  but  a  mere  deduction,  is  not  only  evi- 
dent, but  may  be  more  forcibly  shown  by  means  of 
other  deductions.  Thus,  from  the  former  equation 
we  have 


s  = 


hence,  space  is  the  rate  of  doing  work  per  unit  of 
force  (per  pound). 
Or  again, 


22  KEY  AND  SUPPLEMENT 

hence  velocity  is  the  square  root  of  twice  the  rate  of 
doing  work  per  unit  of  mass  ! 

Or,  again,  in  regard  to  momentum,  page  78, 
Ft  =  Mv  ; 


'   t    ' 

hence,  force  is  the  rate  of  producing  momentum  per 
unit  of  time. 
Or,  again, 

_  M  v 

Tjr. 

hence,  time  is  the  rate  of  producing  momentum  per 
unit  offeree  ! 

Now  all  of  these  are  correct  deductions  ;  but  the 
fundamental  equations  are  established  on  the  hypothe- 
sis that  all,  except  velocity,  are  substantial  quantities. 
The  value  of  J^is  fundamentally  measured  in  pounds. 
Indirectly  it  may  be  measured  in  a  variety  of  ways  as 
shown  above,  and  as  will  be  still  further  shown  in 
Article  86  of  the  text. 

PAGE  17,  ARTS.  31  AND  32.  —  Terrestrial  Gravitation  as 
a  force  causes,  or  tends  to  cause,  bodies  to  move  to- 
wards the  earth,  and  when  a  spring  balance  or  other 
weighing  machine  is  interposed  to  prevent  any  move- 
ment, the  intensity  of  the  impelling  force  may  be  de- 
termined in  pounds  or  an  equivalent. 

PAGE  19,  ART.  35.  —  The  fundamental  idea  of  a  point  of 
application  of  a  force  is  that  of  a  definite  attachment, 
like  the  attachment  of  a  rope,  or  chain,  or  rod  of  iron, 
to  a  body  ;  but  it  is  certain,  in  regard  to  the  forces  of 
nature  —  as  gravity,  chemical  forces,  etc.  —  that  the 
conception  is  erroneous,  for  there  is  no  attachment. 


TO  ELEMENTARY  MECHANICS.  23 

Still  it  may  be  conceived  that  the  force  acts  upon  a 
particle  which  may  still  be  considered  as  the  point  of 
application,  and  thus  the  old  term,  with  its  gross  as- 
sociations, is  useful  in  the  most  refined  sense. 
PAGE  20,  ART.  36. — Inertia  is  a  name  merely  to  express 
the  fact  that  matter  has  not  of  itself  power  to  put 
itself  in  motion,  or  being  in  motion  to  bring  itself  to 
rest,  or  even  to  change  its  rate  of  motion.  Yet  some 
writers  call  Inertia  a  force,  and  others,  with  little  if 
any  more  propriety,  speak  of  the/bra?  of  inertia.  M. 
Morin,  a  French  physicist,  attempted  to  prove  that 
inertia  is  a  force.  He  took  a  prism  standing  on  its 
base,  and  by  a  sudden  pull  or  push  applied  to  its  base 
caused  the  prism  to  fall  backwards.  He  argued  that 
the  falling  over  of  the  prism  indicated  the  action  of  a 
force.  A  force  might  have  been  applied  at  the  top  of 
the  prism  which  would  have  overturned  it  directly, 
and  Morin  argued,  that  when  it  overturned  by  a  sud- 
den action  at  the  base,  there  must  have  been  a  force 
equivalent  to  one  at  the  top,  and  this  he  called  the 
force  of  inertia.  The  fact  is,  any  force  applied  to  a 
body,  not  acting  in  a  line  through  its  centre,  tends  to 
rotate  the  body  ;  and  in  all  cases  where  tho  body  is 
free,  will  cause  it  to  actually  rotate.  If  the  prism 
referred  to,  standing  on  its  base,  be  acted  upon  by  a 
force  applied  at  any  point  above  the  base,  it  will  not 
be  overturned  unless  the  moment  of  the  force  exceeds 
the  moment  of  the  weight  in  reference  to  the  point 
about  which  it  tends  to  turn.  If  the  force  applied  at 
the  base  be  so  intense  as  to  produce  a  rotary  mo- 
ment exceeding  the  moment  of  the  weight  above  re- 
ferred to,  it  will  cause  the  prism  to  fall  by  rotating 


24  KEY  AND  SUPPLEMENT 

backwards ;  but  if  the  force  be  less  intense,  it  will 
simply  cause  the  body  to  slide  on  the  plane. 

Again,  inertia  does  not  fulfill  any  of  the  conditions 
of  a  force.  It  is  not  an  action  between  bodies.  It  is 
not  an  action  in  any  sense.  It  cannot  be  measured 
by  pounds.  It  is  a  negation — an  entire  lack  of  some- 
thing— a  lack  of  force.  We  repeat,  it  is  not  a  force. 

EX  ERCISES. 
PAGK  22. 

1.  The  least  force.  One  object  of  some  of  these 
exercises  is  to  enable  the  student  to  get  a  correct 
idea  of  the  relation  between  force  and  the  result- 
ant motions  of  bodies  by  the  inductive  method. 
The  student  who  has  not  correct  notions  of  these 
relations  will  doubtless  insist  that  it  requires  more 
force  to  move  a  large  body  than  a  small  one ; 
and  he  may  go  so  far  as  to  say  that  it  will  take  10 
pounds  of  pull  or  push  to  move  a  body  weighing  10 
pounds.  But  the  fact  must  be  perceived  that  the 
smallest  force  (10  Ibs.  of  push,  for  instance)  will  just 
as  certainly  move  100  Ibs.  of  matter  free  to  move  in 
the  direction  of  the  push,  as  it  will  one  pound.  It  will 
not  move  the  former  as  rapidly  as  the  latter — or,  in 
other  words,  it  will  not  move  it  as  far  in  the  same 
time.  In  this  way  we  get  an  idea  of  the  fact  that  the 
visible  effect  of  a  force  depends  conjointly  upon  the 
mass  moved,  and  the  space  through  which  it  is  moved 
in  a  given  time.  If  necessary  make  an  experiment 
by  suspending  different  weights  with  equally  long 
strings,  and  pull  them  sidewise  by  a  string  attached 
to  the  body  passing  over  a  pulley— or  edge  of  the 


TO  ELEMENTARY  MECHANICS.  25 

table — and  holding  at  the  suspended  end  a  small 
weight.  It  will  be  found  that  any  weight  which  is 
sufficient  to  overcome  the  friction  of  the  string  on  the 
pulle}7 — or  edge  of  the  table — will  pull  the  heaviest 
weight  sidewise.  Observe  that  the  experiment  is 
simply  to  show  actual  movement,  and  not  the  amount 
of  movement. 

2.  The  least  force.  Also  the  least  force  would  de- 
flect it  from  its  course.  This  shows  that  a  force  will 
have  the  same  effect  upon  a  moving  body  as  upon  one 
at  rest. 

PAGE  23. — 3.  Because  it  is  opposed  by  an  equal  oppo- 
site force. 

4.  100  pounds.     A  man  once  pulled  a  spring  bal- 
ance so  as  to  indicate,  say,  100  pounds.    Another  gen- 
tleman asserted  that  he  could  pull  two  balances  at- 
tached end  to  end  so  that  each  would  indicate  100 
pounds.     This  he  did  to  the  astonishment  of  the  ob- 
servers,  but  their    astonishment   ceased   when   they 
found  that  it  was  no  more  difficult  to  pull  two  in  that 
way  than  one. 

5.  Yes.     The  fact  that  the  boat  is  in  motion,  does 
not  affect  the  result.     Hence  we  have  the   principle 
that  "  action  and  reaction  are  equal  and  opposite." 

6.  No.     To  show  it  (should  there  be  doubt)  assume 
that  a  string  passes  from  each  sled  to  the  hand,  then 
will  the  tension  on  each  be  less  than  10  pounds,  but 
on  both  strings  it  will  be  just  10  Ibs.     Conceive  that 
the  strings  become  one  back  to  the  first  sled — the  ten- 
sion on  the  one  will  be  10  Ibs. — but  on  the  part  back 
of  the    first,  it  will  be  the   same  as   before — which 
was  evidently  less  than  10  Ibs.     If  the  two  sleds  are 


26  KEY  AND  SUPPLEMENT 

of  equal  weight  the  tension  on  the  connecting  cord 
will  be  5  Ibs.  I  have  heard  students  assert  that 
no  tension  could  be  produced  unless  there  were  a  re- 
sisting force — showing  that  they  had  not  yet  a  correct 
conception  of  the  relation  between  forces  and  masses. 
It  requires  force  to  move  a  free  body.  In  such  cases 
I  have  asked  them  to  conceive  that  a  cord  were  at- 
tached to  the  moon,  and  that  they  pulled  upon  it ; 
when  they  will  severally  admit  that  a  pull  of  ten  or 
more  pounds  may  be  easily  exerted.  If  the  sleds 
were  not  very  heavy,  it  would  not  be  possible  for  the 
boy  to  run  sufficiently  fast  to  maintain  a  constant  pull 
of  10  pounds  for  a  long  distance ;  but  if  it  can  be 
done  for  a  few  feet  only,  it  will  answer  the  purpose 
of  the  illustration. 

T.  No  tension. 

PAGE  23,  ART.  48. — The  so-called  Three  Laws  of  Motion 
did  not  spring  suddenly  into  philosophy.  There  was 
a  long  period  of  darkness  succeeded  by  twilight  and 
dawn  before  the  truth  shone  out  clearly.  The  prin- 
ciples of  the  three  laws  were  known,  and  to  a  consid- 
erable extent  realized,  before  Newton's  time,  but  per- 
haps they  had  not  been  so  clearly  and  sharply  defined 
by  any  preceding  writer,  and  much  less  had  they  been 
made  the  foundation  of  mechanical  science ;  hence 
there  is  a  certain  propriety  in  calling  them  Newton's 
Laws.  Correct  notions  in  regard  to  these  principles 
date  from  the  time  of  Galileo.  Prior  to  his  day,  the 
leading  philosophy  was  Aristotelian.  Aristotle  flour- 
ished between  300  and  400  years  before  Christ.  In 
his  philosophy  there  was  no  distinct  idea  of /0m?  as  a 
cause,  much  less  any  idea  of  a  relation  between  the 


TO  ELEMENTARY  MECHANICS.  27 

cause  of  motion  and  the  momentum  produced.  He 
taught  that  a  heavy  body  would  fall  faster  than  a 
lighter  one — that  when  a  body  is  thrown  by  the  hand 
it  ou<^ht  to  cease  to  move  as  soon  as  it  left  the  hand 

o 

were  there  no  surrounding  impulses,  but  that  it  con- 
tinued to  move  because  the  hand  sets  in  motion  the 
air  about  the  body,  and  that  the  air  acted  afterwards 
in  impelling  the  body.  He  divided  motions  into 
Natural  and  Violent ;  the  former  of  which  is  illustra- 
ted by  a  falling  body,  in  which  the  motion  is  constantly 
increasing,  and  the  latter  by  a  body  moving  on  the 
ground,  where  the  motion  is  constantly  decreasing. 
It  must  not,  however,  be  inferred  that  philosophers 
had  no  idea  of  cause  and  effect.  Some  general  notions 
of  this  kind  have  always  been  entertained. 

Between  the  period  of  Aristotle  and  Galileo,  many 
important  principles  were  established.  Archimedes 
(born  287  B.  c.)  developed  some  important  properties 
of  the  Centre  of  Gravity,  established  the  principle  of 
the  Lever,  and  some  of  the  principles  of  Hydrostatics. 
History  seems  to  show  that  the  advanced  position  se- 
cured by  this  eminent  philosopher  was  not  main- 
tained, and  that  little  or  no  advance  was  made  until 
the  time  of  Stevin — or  Stevainus,  as  commonly  writ- 
ten (1548-1620).  His  determination  of  the  condi- 
tions of  equilibrium  of  the  inclined  plane  is  so  in- 
genious, it  is  worth  repeating.  Consider  two  inclined 
planes  having  a  common  vertex  and  horizontal  base. 
Conceive  a  uniform  long  chain  to  be  placed  on  them, 
and  joined  underneath  so  as  to  hang  freely.  He 
showed  that  it  would  hang  at  rest  without  friction, 
because  any  motion  would  only  bring  it  into  the  same 


28  KEY  AND  SUPPLEMENT 

condition  in  which  it  was  at  first.  The  part  hanging 
below  would  evidently  be  in  equilibrium  by  itself, 
hence  if  that  part  be  cut  off  the  remaining  part  will 
be  at  rest ;  hence  the  condition  of  equilibrium  is — 
the  weights  on  each  part  must  l)e  exactly  proportional 
to  the  lengths  of  the  planes.  If  one  side  becomes  verti- 
cal the  same  proportion  holds  true. 

Galileo  forms  the  grand  connecting  link  between 
the  philosophers  of  the  ancient  and  modern  physical 
sciences.  He  was  born  at  Pisa,  February  18th,  1564, 
39  years  before  the  death  of  Michael  Angelo,  and  21 
years  after  the  death  of  Copernicus,  and  died  on  the 
8th  of  January,  1642,  the  year  in  which  Sir  Isaac 
Newton  was  born.  The  science  of  motion  began  with 
him.  He  taught  that  motion  was  due  to  force — that 
all  bodies  in  a  vacuum  would  fall  with  equal  velocities 
— that  inertia  of  matter  implies  persistence  of  condi- 
tion— he  gave  a  satisfactory  definition  to  momentum 
— also  stated  with  approximate  precision  the  princi- 
ple that  "  action  and  reaction  are  equal  " — also  estab- 
lished the  principle  of  "  virtual  velocities,"  which 
was  made  by  Lagrange  to  include  all  of  mechanical 
science  in  one  expression.  He  made  a  mathematical 
analysis  of  the  strength  of  beams,  of  projectiles,  of 
the  pendulum,  of  floating  bodies,  and  of  the  inclined 
plane.  For  his  investigations  in  other  fields  of  sci- 
ence— see  some  biographical  sketch. 
PAGE  23,  ART.  49. — First  Law  of  Motion.  Says  "Wliew- 
ell,  in  his  History  of  the  Inductive  Sciences :  "It 
may  be  difficult  to  point  out  who  first  announced  this 
Law  in  a  general  form."  We  have  already  seen  that 
the  facts  involved  in  it  were  recognized  by  Galileo. 


TO  ELEMENTARY  MECHANICS.  29 

It  is  equivalent  to  saying  that  every  change  is  due  to 
a  cause,  and  yet,  to  cover  the  entire  ground,  this  state- 
ment needs  modification.  Motion  is  due  to  a  cause, 
but  when  the  cause  ceases,  the  motion  of  a  free  body 
does  not  cease,  it  simply  becomes  uniform.  Change 
of  position  is  not  then  necessarily  due  to  a  coexisting 
cause,  but  may  be  due  to  a  cause  remote  in  time. 
Change  of  condition,  however,  requires  a  present  act- 
ing cause  ;  and  the  latter  will  produce  either  a  change 
in  the  rate  of  motion,  or  of  the  direction  of  motion  or 
of  both. 

The  law  cannot  be  proved  by  direct  experiment, 
for  it  is  practically  impossible  to  remove  from  the 
body  all  acting  forces,  and  hence  uniform  motion  un- 
der the  action  of  no  forces  is  not  realized  by  experi- 
ment. It  may,  however,  be  observed  that  the  less  the 
resistance  the  more  nearly  uniform  will  be  the  motion, 
and  hence  we  are  led  to  infer  that  if  all  resistance 
could  be  removed,  the  motion  would  be  strictly  uni- 
form. Similarly,  it  is  observed  that  a  body  projected 
on  a  very  smooth  plane  moves  so  nearly  in  a  straight 
line  that  we  are  led  to  infer  that  if  there  were  no  de- 
flecting causes  the  path  would  be  exactly  straight. 
The  law  is  the  result  of  induction  rather  than  of 
proof,  but  it  appears  so  perfectly  reasonable  that  we 
assent  to  it  as  soon  as  it  is  properly  illustrated.  The 
strongest  proof  of  its  correctness  lies  in  the  fact  that 
deductions  founded  on  this  hypothesis  agree  with 
the  results  of  observation. 

The  process  of  induction  consists  in  conceiving 
clearly  the  law,  and  in  perceiving  the  subordination 
of  facts  to  it. 


30  KEY  AND  SUPPLEMENT 

PAGE  23,  ART.  50. — Second  Law.  Change  of  motion  is 
in  proportion  to  tJie  acting  force.  If  all  bodies  were 
of  equal  size  it  would  only  be  necessary  to  consider 
their  relative  velocities  in  determining  the  effect  of 
forces.  But  a  larger  body  requires  more  force  than  a 
smaller  one  of  the  same  substance  to  produce  the  same 
velocity  under  the  same  circumstances  ;  in  other  words 
— mass  and  velocity  are  both  involved — and  the  term 
motion  here  means  momentum.  This  also  agrees 
with  Newton's  explanation  of  the  term.  It  is  better, 
therefore,  to  word  the  law  thus :  Change  of  momen- 
tum is  in  proportion  to  the  acting  force.  It  must  be 
particularly  observed  that  the  law  does  not  assert  that 
momentum  varies  as  the  force  y  but  that  it  is  a  change 
of  the  momentum  that  varies  as  the  force. 

This  law  was  clearly  perceived  by  Galileo,  and  by 
means  of  it  and  the  first  law,  he  determined  that  the 
path  of  a  projectile  in  a  vacuum  was  a  parabola.  The 
law,  however,  was  not  considered  fully  established 
until  the  theory  in  regard  to  the  motion  of  the  earth, 
involving  both  this  law  and  the  law  of  universal  grav- 
itation, was  realized.  The  triumph  of  both  laws  was 
complete  at  the  same  time. 

PAGE  24,  ART.  51. — Third  Law.  Action  and  Reaction 
are  equal  and  opposite.  The  first  and  second  laws 
refer  to  one  body  only  ;  this  involves  two  bodies.  It 
asserts  that  an  action  between  two  bodies  is  of  the 
same  intensity  upon  each,  but  that  the  direction  of 
action  upon  one  is  directly  opposed  to  that  upon  the 
other.  Every  action  implies  an  equal  opposite  action 
— one  being  called  a  reaction  in  reference  to  the 
other.  No  force,  acting  in  one  direction  only,  exists  ; 


TO  ELEMENTARY  MECHANICS.  31 

it  is  always  accompanied  by  an  equal  but  opposite  ac- 
tion. No  force  is  known  to  exist  without  the  pres- 
ence of  matter.  Force  does  not  act  in  curved  lines  ; 
the  action  and  reaction  between  two  particles  is  in 
the  right  line  adjoining  them. 

Newton  gave  three  examples  illustrating  this  law : 

If  any  one  presses  a  stone  with  his  finger,  his  fin- 
ger is  also  pressed  by  the  stone. 

If  a  horse  draws  a  stone  fastened  to  a  rope,  the' 
horse  is  drawn  backward,  so  to  speak,  equally  towards 
the  stone. 

If  one  body  impinges  on  another,  changing  the 
motion  of  the  other  body,  its  own  motion  experiences 
an  equal  change  in  the  opposite  direction. 

It  does  not  seem  rational  that  the  stone  will  pnsli 
the  finger  in  the  same  sense  that  the  finger  presses 
the  stone.  The  finger  appears  to  be  an  active  agent 
while  the  stone  is  inert.  In  the  strictest  sense  we 
should  say  that,  in  the  attempt  to  press  a  stone  a 
force  is  developed  between  the  finger  and  the  stone, 
which  force  acts  equally  in  opposite  directions ;  in 
one  direction  against  the  finger,  in  the  other  against 
the  stone. 

Similarly,  in  regard  to  the  horse  and  stone.  In  the 
former  example  the  condition  is  statical,  but  in  this 
the  horse  is  supposed  to  move  the  stone.  The  horse, 
evidently,  is  not  actually  pulled  backward,  although 
there  is  an  actual  backward  pull  upon  the  horse  by 
the  rope.  The  fact  is,  that,  in  the  effort  to  draw  the 
stone,  a  force  is  developed  which  produces  tension  in 
the  rope,  which  tension  acts  to  pull  the  stone  one 
way,  and  the  horse  the  opposite  way.  As  the  horse 


32  KEY  AND  SUPPLEMENT 

is  able  to  take  a  footing  on  the  earth,  he  is  able  to 
exert  a  force  on  the  rope  equal  to,  or  exceeding,  that 
necessary  to  overcome  the  friction  of  the  stone,  and 
thus  move  the  stone.  The  pressure  between  the  horse 
and  the  earth  also  acts  in  opposite  directions. 

In  the  third  illustration  another  idea  is  presented. 
Motion  is  used  in  the  sense  of  momentum,  as  before 
stated,  and  it  should  read,  the  change  of  momentum 
in  two  impinging  bodies  is  the  same  in  both  bodies 
but  in  opposite  directions.  Tin's  is  a  necessary  result 
of  the  second  and  third  laws.  The  forces  bcinsj 

o 

measured  by  the  change  in  the  momentum,  and  the 
pressures  being  equal  between  the  impinging  bodies, 
it  follows  that  the  changes  in  their  momenta  must  be 
equal.  Hence  if  one  loses  momentum  the  other  must 
gain,  and  the  loss  in  one  case  must  equal  the  gain  in 
the  other. 

Thus  if  the  body  whose  mass  is  J!/i  impinges  upon 
another  whose  mass  is  Jf2,  Vi  and  vz,  their  respect- 
ive velocities  before  impact,  and  v\  and  v'z  their  re- 
spective velocities  at  any  instant  after  the  first  contact, 
and  assuming,  as  we  will  in  establishing  the  formula, 
that  the  velocities  are  in  the  same  direction,  in  which 
case  it  will  only  be  necessary  to  change  the  sign  of  one 
of  the  velocities  if  they  move  in  opposite  directions  ; 
then  will  the  momentum  of  MI  before  impact  be 


and  after  impact 

and  as  M^  is  supposed  to  be  the  impinging  body  it 
will  lose  velocity,  and  we  have  for  the  momentum 
lost  by  MI 


TO  ELEMENTARY  MECHANICS.  33 


Similarly,  the  momentum  gained  by  Mz  will  be 


which,  according  to  the  third  law,  must  equal  that  lost 
by  the  former  body  ;  hence  for  all  stages  of  the  mo- 
tion after  the  first  contact,  not  only  during  compres- 
sion, but  also  at  and  after  separation,  we  have 


If  J/2  be  the  impinging  body,  we  have 


which  easily  reduces  to  the  former  equation.  If  they 
move  in  opposite  directions  before  impact  make  i\  or 
v2  negative,  the  impact  is  here  supposed  to  be  direct 
and  central,  for  which  case  the  equations  are  true 
whether  the  bodies  be  elastic  or  non-elastic.  Several 
cases  are  discussed  on  pp.  85-90  of  the  text. 

(It  appears  that  Whewell,  in  his  History  of  the  Inductive 
Sciences,  has  not  drawn  a  sufficiently  clear  distinction  between 
the  second  and  third  laws.  He  appears  to  hold  that  the  third 
law  gives  a  measure  of  the  pressure  or  force,  whereas  the  sec- 
ond law  is  the  only  one  of  the  three  that  gives  it). 

PAGE  24,  ART.  52.  —  We  are  not  informed  who  first 
gave  the  laws  for  the  composition  and  resolution  of 
forces  ;  but  Galileo  was  one  of  the  first  to  make  use 
of  them  in  explaining  curvilinear  motions.  The 
method  was  systematized  by  Descartes  by  the  aid  of 
the  systems  of  rectilinear  axes. 


KEY  AND  SUPPLEMENT 


EXERCISES. 
PAGE  26. 

1.  Because  the  force  of  gravitation  draws  it  from 

the  rectilineal  path  in  which  it  was  projected. 

2.  The  least  force. 

3.  500  pounds. 

PAGE  27. 

4.  Midway  between  their  initial  positions. 

5.  He  must  aim  to  walk  southwesterly.    To  find  the 

direction,  draw  a  line,  AB,  of  any  A B 

length  to  represent  the  one  mile 
due  east,  and  a  line,  AC,  perpendic- 
ular to  AB,  and  of  such  length 
that  the  hypothenuse,  BC,  will  be 
three  times  as  long  as  AB  /  then 
will  the  angle  AB  C  represent  the 
required  direction. 
We  have 

cos  B  =  - ; 

.-.  ABC  =  40°  32°. 

6.  Make  AB  =  3,  the 

angle  DBG  =  45°, 
and  BC  equal  to 
8;  join  AC,  then 
will  AC  represent 
the  resultant  direc- 
tion. A  numerical 
solution  gives  DA  C 
=  33°  10'  and  the 
course  will  be  S.  56°  50'  E. 


TO  ELEMENTARY  MECHANICS.  35 

BD  =  DC  =  8  sin  45°  =  8  x  4  A/2  =  4  V2 
^4j9  =  3  +  4  V2 

n^         4V1           32  - 12  A/2 
tan.  .ZM  C  = 7=  = — -  = 

3  +  4  V2  23 

0-6535;  .-.  DAC=  33°  10'. 

The  velocity  will  be  10-34  miles  per  hour. 

7.  It  will  be   10  x  cos  45°  =  $  A/2  x    10  =  5  x 

1.4142  +  =7.0710+. 

8.  Yes,  and  will  move  towards  the  rear  end  of  the 

car  (First  Law).  Because  it  tends  to  preserve 
the  velocity  which  it  had  just  before  the  col- 
lision. 

9.  In  the  first  edition  Fz  was  30  Ibs.     It  was  in- 

tended to  be  20  Ibs.  so  as  to  show  that  the  re- 
sultant of  two  velocities  may  be  the  same  as 
those  producing  it.  In  this  case  the  triangle 
of  velocities  will  be  equilateral,  and  hence 
each  of  the  angles  will  be  60°. 

If  20  and  30  pounds  be  used,  we  have  for 
the  diagonal  of  the  parallelogram 


,#  =   A/900  +  400+2  x  20  x  30  x  0-5 
=  43-59  Ibs. ; 

and  for  the  angles  36°  35'  12",  and  23°  24'  48" 
respectively. 

PAGE  27,   ARTS.  59,  60. — The  intensity  of  a  force  is 
known  only  by  its  results. 

PAGE  29. 

ART.  63. — The  case  of  a  force  acting  normal  to  the 
path  of  a  body  was  a  difficult  one  with  the  philoso- 


36  KEY  AND  SUPPLEMENT 

phers  living  about  Galileo's  time,  and  is  not  the 
easiest  to  explain  by  elementary  processes  at  the 
present  time.  It  will  be  considered  in  Chap,  xvi., 
p.  220  of  the  text. 

ART.  64. — We  say  that  gravity  tends  to  draw  bodies 
towards  each  other ;  but  we  only  know  that  it 
causes  them  to  move  towards  each  other  when  free. 
It  is  quite  as  proper  to  speak  of  its  pushing  as  of 
drawing  them  towards  each  other. 

Some  attempts  have  been  made  to  explain  the 
essential  nature — or  the  cause — of  gravitation.  One 
of  the  most  celebrated  of  these  theories  was  given 
by  one  Le  Sage.  According  to  his  theory,  lines  of 
force  acted  in  all  conceivable  directions  through 
space,  and  as  two  bodies  intercept  those  lines  which 
would  pass  through  both  bodies,  there  would  be 
more  force  to  drive  them  towards  than  from  each, 
other.  (See  Theories  of  Gravitation  by  "Win.  JB. 
Taylor,  of  the  Smithsonian  Institute,  Washington, 
D.  C.)  But  no  theory  thus  far  suggested  is  con- 
sidered sufficient  to  account  for  the  fact  of  gravi- 
tation. 

PAGE  29,  ART.  65. — The  law  of  universal  gravitation  is 
one  of  the  discoveries  which  aided  in  immortalizing 
the  name  of  Sir  Isaac  Newton.  He  first  conceived  the 
nature  of  the  law,  and  then  proceeded  to  prove  it.  He 
assumed  that  if  the  law  was  correct  it  ought  to  explain 
the  circular  motion  of  the  moon  about  the  earth — in 
other  words,  that  the  pulling  force  (so  to  speak)  of 
gravity  at  the  moon  would  be  just  sufficient  to  draw 
it  the  required  amount,  from  a  tangent  to  the  orbit 


TO  ELEMENTARY  MECHANICS.  37 

If  g  be  the  acceleration  of  a  pulling  body  at  the  sur- 
face of  the  earth,  at  the  moon  it  will  be  g  ~-  D2 — 
where  D  is  the  number  of  diameters  of  the  earth  be- 
tween the  center  of  the  earth  and  center  of  the  moon, 
and  is  about  60-3612 ;  and  as  g  =  32-246  ft.,  we 
would  have  at  the  moon,  the  acceleration  32-246  -4- 
(60-3612)2  =  0-0088  +  .  The  force  at  the  moon 
which  would  produce  this  acceleration  equals  the  cen- 
trifugal force,  and  is  given  by  the  last  equation  of 
Art.  315  of  the  text,  and  is 

4**p 

Force  =  m  -=- 


where  r  is  the  radius  of  the  orbit,  T  the  time  of  a 
complete  revolution,  and  m  the  mass  of  the  body. 
But  the  force  divided  by  the  mass  equals  the  accel- 
eration— as  is  shown  by  the  last  of  the  equations  of 
Article  86 ; 

7      , .  Force       4:7tzr 

.*.  accelerat^on  =  =  —f^- , 

in          T* 

which  applied  to  the  moon,  and  reduced,  gives  0-0089 
+  (see  Art.  319).  The  two  results  should  agree  if  the 
law  and  data  are  both  correct.  It  will  be  seen  that 
the  value  of  the  radius  of  the  earth  enters  into  the 
computation,  the  correct  value  of  which  was  not  known 
to  Newton  at  the  time  of  his  first  investigations. 
Some  fifteen  years  after  he  began  the  investigation, 
while  attending  a  lecture  in  London,  he  obtained  the 
correct  value  of  the  radius,  with  which  he  proved  the 
truth  of  his  proposition. 


38  KEY  AND  SUPPLEMENT 

The  analysis  by  which  the  above  result  is  reached 
is  anticipated,  but  it  will  enable  the  reader  to  under- 
stand why  an  error  in  the  true  value  of  the  radius  of 
the  earth  vitiated  the  first  result. 

It  is  questionable  whether  the  story — that  Newton 
conceived  the  law  of  gravitation  by  seeing  an  apple 
fall  from  a  tree — so  often  taught  to  juveniles,  is  not 
purely  fictitious.  It  is  certain  that  he  did  not  con- 
sider the  law  established  for  fifteen  years  after  he 
first  conceived  it,  during  which  time,  it  is  said,  he  re- 
viewed his  work  many  times. 

PAGE  30,  ART.  67. — A  history  of  pendulum  experiments 
would  furnish  material  for  a  book.  The  'mathematical 
pendulum  is  an  ideality,  but  a  very  useful  one  in  dis- 
cussing the  subject.  Compound  pendulums  are 
necessarily  used  in  making  experiments.  The  most 
practical  method  of  determining  the  acceleration  due 
to  gravity  is  by  means  of  a  pendulum  ;  and  some  of 
the  results  thus  found  are  given  on  p.  244  of  the  text. 
The  length  of  the  seconds'  pendulum  has  also  been 
used  for  determining  the  standard  of  linear  measure. 
Thus,  the  English  law  requires  that  the  length  of  the 
yard  shall  be  to  the  length  of  the  simple  pendulum 
vibrating  seconds  at  the  Tower  of  London  reduced 
to  the  level  of  the  sea  as  36  to  39-13908.  (See  Art. 
328.) 

PAGE  31,  AET.  68.— The  formula 

g  =  32-1726  -  0-08238  cos.  2Z, 

is  given  in  The  Mecanique  Celeste,  Tome  iii.  v.,  §  42, 
[2,049]. 


TO  ELEMENTARY  MECHANICS.  39 

EXAMPLES. 

PAGE  35. 

,08  _ 

1.  h  =  2j  .'.  v  =  V2gh  =  V2  x  32|  x   100  = 
80-20  feet. 


v      300 
2.  rf  =  -  =  77P—  =  9-3  seconds. 

ff     S2i 


3.  t  =       =  3-109  seconds. 
9 


h  =  v0t-  $g?  =  100  x  3409  -  $  x  32£  x  (3-109)2  = 
1554  feet. 

PAGE  36. 

4.  Acceleration  =  32|  ft.  per  sec. 

x  60  =  1,930  ft.  per  minute. 


5.  Acceleration  per  second  =  32-16666  feet,  =  32- 

16666  -4-  3-28  =  9-807  +  metres.     For  4  sec- 
onds, 9-807  x  4  =  39-228  metres. 

6.  h  =  lfffi  +  v0t,  (Eq.  (6)  p.  34.); 


.-.  120  =  |  x        -  fl  +  25*. 

D 


Solving  for  t  we  find 

t  =  2-0627  +  sec. 
Also  (Eq.  (11)  p.  13.) 

v  =v0  +gt 


40  KEV  AND  SUPPLEMENT 


7. 


=  25  +  x  2-0627 

D 

=  91-35  feet. 
iqq 


-       15°  A    i/ 
—  ~  ^6  +  V 


1800  x  193  +  22500 


193  1932 

-150+608-19 

193 
=  2-37  seconds. 

8.  For  the  falling  body  (Eq.  (2)  p.  34) 


and  for  the  body  projected  (Eq.  (8)  p.  34) 

h-vt-  \g$. 
Eliminating  h  gives 

v  =  gt; 
but 

<  =  ^; 

V 

hence  eliminating  t  gives 


9.  The  sound  will  be  -~^r   seconds    in   returning; 


hence  the  time  of  falling  will  be 


TO  ELEMENTARY  MECHANICS.  41 

h 
'  1130  ~ 


or 


.*.  A  =  331-6+  feet 

10.  Let  x  =  the  distance  upward  from  the  lower 
point  to  the  point  of  meeting  ;  then  will  the 
point  of  meeting  be  a  —  x  from  the  upper 
point.  If  t  be  the  time  of  meeting,  we  have 
for  the  falling  body 


a  —  x  =  vt  + 
and  for  the  body  projected  upward 

x=  Vt-\(j&. 
Adding  we  have 

a=(V+v)t; 


V+  v' 
which  in  the  second  equation  gives 


42  KEY  AND  SUPPLEMENT 

The  distance  below  the  highest  point  will  be 

a>       f  TT-  0     \ 

a  —  x=  -77 (  V  +  lq  -TT )• 

V  +  v  \  V  +  vj 

EXERCISES. 

PAGE  36. — 1.  He  will.  To  draw  it  at  a  uniform  veloc- 
ity he  must  overcome  the  friction  only,  but  at 
an  increasing  velocity  he  not  only  overcomes 
the  friction  but  also  exerts  additional  force  to 
overcome  the  inertia  of  the  mass  (see  Second 
Law). 

2.  Attraction  is  more. 

3.  Because  the  air  resists  less. 

4.  Bounce. 

PAGE  37. 

5.  10  inches. 

6.  More,  for  the  force  of  gravity  is  less  at  the  equa- 

tor than  at  the  poles. 

7.  It  will  neither  gain  nor  lose  in  weight  if  weighed 

with  the  same  beam  scales.  It  will  lose  in 
weight  if  weighed  with  the  same  spring  bal- 
ance, for  the  resistance  of  the  spring  will  re- 
main constant,  while  the  force  of  gravity  will 
be  less. 

8.  9-807  +  metres  per  second. 

PAGE  38,  ART.  78. — In  the  last  two  lines  of  the  page 
it  is  assumed  that  the  attraction  of  a  sphere  upon  an 
external  particle  varies  inversely  as  the  square  of  the 
distance  from  the  center  of  the  sphere.  We  here 
submit  a  proof  of  the  truth  of  the  statement.  New- 


TO  ELEMENTARY  MECHANICS.  43 

ton,  in  his  Principia,  proved  the  proposition  of  the 
statement  geometrically  ;  we  now  use  the  calculus. 

Let  ABD  be  a  spherical  shell,  center  (7,  radius  a, 
P  the  posi- 
tion of  an  ex- 
ternal parti- 
cle.    C  o  n- 
c  e  i  v  e  two 
consecutive 
radial    lines 

drawn  from   P,  cutting  the  shell   in   the  points  A 
and  B. 

Proof.     Let  ds  be  an  element  of  length  of  the  circle  at  A, 
PA  =  r,  PC  =  c,   CB  =  a,  k  =  thickness  of  shell,  8  =  den- 
sity.    Conceive  a  line  joining  Pand  C,  Q  =  APC,  andy  =  per- 
pendicular from  A  upon  PC. 
Then 

2  Ttykds  =  volume  of  shell  generated  hy  the 

revolution  of  A  about  PC, 
2  nySkdii  =  mass  thus  generated  . 

The  attraction  upon  the  particle  will  be 

2  ndkyds 
—  p        • 

which  resolved  along  the  axis  PC  gives 
2  itSkyds 


Let  p  =  CE  =•  the  perpendicular  from  C  upon  PB,  then 

p  =  c  sin  6  ;  .  •.  dp  =  e  cos  Q  d  Q  ; 
also  r*  —  2rc  cos  0  +  c*  =  a*  ; 

dr  re  sin  9 


and 


'  dO  r  —  e  cos 

ds  ar 


dO       r  —  c  cos  6  ' 
(for  ds*  =  dr*  +  rW). 


44  KEY  AND   SUPPLEMENT 

Hence 
2  itSkyds  2  TtSky  cos  0          ardO 

5 COS  O   =    5 .  — 

r  r  r  —  c  cos  0 

2  itSkyadp 
~  cr  (r  —  c  cos  6) 

x       _pdp 


Ya*  _  p* 

which  gives  the  attraction   for  a  circular  element  of  the  shell ; 
hence  for  the  entire  shell  we  have 

pdp 


4:Tt8Jca_    f 
-c*-   J 


hence  for  a  constant  radius  a,  the  attraction  varies  inversely  as 
the  square  of  the  distance  of  the  particle  from  the  center  of  the 
shell,  which  was  to  be  proved. 

If  c  =  a,  we  have 

hence  the  attraction  of  a  spherical  shell  upon  a  particle  in  its 
surface  is  independent  of  the  dimensions  of  the  shell. 

To  find  the  attraction  of  a  homogeneous  sphere  upon  an  exter- 
nal particle,  make  k  =  da  and  integrate,  and  we  have 

4  itS     fa    , ,         4     xSa3 
(fda  = 3 — 


r 

Jo 


S 

=  volume  X  -5  . 
c 


PAGE  39. 

ART.  79.  —  Weight,  to  an  uneducated  person,  is  con- 
ceived to  be  essential  to  matter.  Such  an  one  would 
doubtless  assert  that  a  body  falls  to  the  earth  because 
it  is  heavy  ;  but  to  the  student  of  mechanics,  weight 


TO  ELEMENTARY   MECHANICS.  45 

is  nothing  but  the  measure  of  a  force,  the  magnitude 
of  which  depends  upon  the  quantity  of  matter  con- 
stituting the  body. 

PAGE  41. 

ART.  83. — The  unit  of  mass  might  be  the  piece  of 
platinum  which  is  used  as  the  standard  pound  (see 
Art.  33),  but  as  we  have  occasion  to  compare  the 
force  of  gravity  at  different  places,  and  as  the  force 
of  gravity  at  London  is  assumed  to  be  32-J-  feet,  we 
have  chosen  to  consider  the  unit  of  mass  as  about 

—  of  the  weight  at  that  place. 
62% 

PAGE  42. 

ART.  85. — Density  is  sometimes  used  in  the  sense 
of  specific  gravity,  and  if  the  density  of  water  be 
taken  as  unity,  the  specific  gravity  of  any  substance 
(compared  with  water)  will  equal  its  density.  We 
prefer  to  make  the  definition  conform  to  the  sense 
used  in  mechanics.  The  following  are  the  definitions 
given  by  several  authors  : 

By  the  density  of  a  body  is  meant  its  mass  or  quantity  of  mat- 
ter compared  with  the  mass  or  quantity  of  matter  of  an  equal  vol- 
ume of  some  standard  body  arbitrarily  chosen. — TOWNE'S  Ele- 
mentary Chem.,  p.  29. 

Density  is  a  term  employed  to  denote  the  degree  of  proximity 
of  the  atoms  of  a  body.  Its  measure  is  the  ratio  arising  from 
dividing  the  number  of  atoms  the  body  contains  by  the  number 
contained  in  an  equal  volume  of  some  standard  substance  whose 
density  is  assumed  as  unity.  The  standard  substance  usually 
taken  is  distilled  water  at  the  temperature  of  38°. 75  F. — BART- 
LETT'S  Elements  of  Analytical  Mechanic*,  p.  29. 

Enfin  si  1'on  represente  par  Z>  la  masse,  sous  1'unite  de  volume 
du  corps  que  Ton  considere,  D  sera  ce  qu'on  nomine  la  Densite 


46  KEY  AND  SUPPLEMENT 

de  ce  corps.  On  prend  commandment  pour  unite  de  densite  celle 
de  1'eau  destillee  a  cette  derniere  temperature  (4°  du  thennome- 
tre  centigrade).— POISSON,  Traite  de  Mecanique,  page  108. 

The  quantity  of  matter  in  a  body  does  not  depend  on  the  size 
of  the  body  only,  but  also  on  the  closeness  with  which  the  par- 
ticles are  packed.  This  difference  is  denned  as  a  difference  of 
density.  Thus  there  is  more  matter  in  a  cubic  inch  of  lead  than 
in  a  cubic  inch  of  oak,  and  this  is  expressed  by  saying  that  the 
density  of  lead  is  greater  than  the  density  of  oak. — MAGNUS,  Les- 
sons in  Elementary  Mechanics,  p.  60. 

Experiment  shows  that  the  weight  of  a  certain  volume  of  one 
substance  is  not  necessarily  the  same  as  the  weight  of  an  equal 
volume  of  another  substance.  Thus  seven  cubic  inches  of  iron 
weigh  about  as  much  as  five  cubic  inches  of  lead. 

We  say  then  that  lead  is  denser  than  iron,  and  we  adopt  the 
following  definitions.  When  the  weight  of  any  portion  of  a  body 
is  proportional  to  the  volume  of  that  portion,  the  body  is  said 
to  be  of  uniform  density.  And  the  densities  of  two  bodies  of 
uniform  density  are  proportional  to  the  weights  of  equal  volumes 
of  the  bodies.— TODHUNTER,  Mechanics  for  Beginners,  p.  7. 

The  density  of  a  body  is  the  mass  comprised  under  a  unit  of 
volume. — SILLIMAN,  First  Principles  of  Philosophy,  p.  67. 

Density  "1.3  the  quantity  of  matter  contained  in  a  unit  volume  ; 
the  absolut-j  density  or  the  closeness  with  which  the  particles  are 
packed  being  uniform  throughout  that  unit  volume.  This  def- 
inition is  directly  applicable  if  a  body  is  homogeneous  ;  but  if  it 
is  heterogeneous,  and  the  density  varies  from  point  to  point,  the 
density  at  any  point  is  the  quantity  of  matter  contained  in  a 
unit  volume  throughout  which  the  density  is  the  same  as  that  at 
the  point.  Density  is  usually  measured  by  means  of  comparison 
with  some  substance  the  density  of  which  is  assumed  to  be  the 
unit-density. — PRICE,  Infinitesimal  Calculus,  p.  164. 

The  quantity  of  matter  in  a  body,  or  as  we  now  call  it,  the 
mass  of  a  body,  is  proportional,  according  to  Newton,  to  the  vol- 
ume and  the  density  conjointly.  In  reality  the  definition  gives 
us  the  meaning  of  density  rather  than  of  mass,  for  it  shows  us 


TO  ELEMENTARY  MECHANICS.  47 

that  if  twice  the  original  quantity  of  matter,  air  for  example,  be 
forced  into  a  vessel  of  given  capacity,  the  density  will  be  doubled 
and  so  on.  But  it  also  shows  us  that,  of  matter  of  uniform  den- 
sity, the  mass  or  quantity  is  proportional  to  the  volume  or  space 
it  occupies. — THOMSON  AND  TAIT,  Treatise  on  Natural  Philoso- 
phy, p.  162. 

Heaviness  (Fr.,  densite,  Ger..  dichtigkeit)  is  the  intensity  with 
which  matter  fills  space.  The  heavier  a  body  is,  the  more  mat- 
ter is  contained  in  the  space  it  occupies.  The  natural  measure 
of  heaviness  is  that  quantity  of  matter  (the  mass)  which  fills  the 
unity  of  volume;  but  since  matter  can  only  be  measured  by 
weight,  the  weight  of  a  unit  volume,  e.  g.  of  a  cubic  meter  or 
of  a  cubic  foot  of  another  matter,  must  be  employed  as  a  meas- 
ure of  its  heaviness. 

The  product  of  the  volume  and  the  heaviness  is  the  weight. 

The  heaviness  of  a  body  is  uniform  or  variable  according  as 
equal  portions  of  the  volume  have  equal  or  different  weights. — 
WEISBACH,  Mechanics  of  Engineering,  p.  160. 

The  density  of  a,  body  is  the  degree  of  closeness  between  its 
particles.  The  term  depends  upon  the  hypothesis  that  the  ulti- 
mate particles  of  matter  have  weight,  and  therefore  mass  pro- 
portional to  their  bulk.  It  coincides  with  specific  gravity. — 
PBOP.  NICHOL,  Cyclopcedia  of  the  Physical  Sciences,  page  177. 

On  sait  que  le  poids  d'un  corps  varie  avec  1'intensite  de  la 
pesanteurmaisque  sa  masse  ne  varie  pas.  Sous  1'influence  de  la 
meme  pesanteur,  par  exemple  en  un  meme  lieu  du  globe,  le 
poids  est  evidemment  proportionnel  a  la  masse  et  le  rapport  des 
poids  de  deux  substances  sous  le  meme  volume  sera  precisement 
celui  de  leurs  masses  sous  le  meme  volume  ;  de  la,  la  synonymie, 
qui  existe  entre  les  mots  poids  specifique  et  densite,  qui  ex- 
prime  ses  rapports. — WUBTZ,  Dictionnaire  de  Chimie.  Sous  Den- 
eite. 

The  density  of  a  body  is  the  ratio  of  its  mass  to  its  volume.— 
SMITH,  Elementary  Treatise  of  Meclianics,  p.  44. 


48  KEY  AND  SUPPLEMENT 

EXERCISES. 

PAGE  43. 

1.  The  matter  outside  of  one-half  the  radius  would 

produce  no  effect ;  and  that  within  would  at- 
tract as  if  it  were  all  at  the  centre.  The  sphere 
of  one-half  the  radius  will  contain  one-eighth 
the  matter,  and  the  inverse  square  of  the  dis- 
tance will  be  4  ;  hence  the  weight  will  be 
£  x  4  x  10  =  5  Ibs.  We  get  this  result  more 
directly  by  saying,  as  in  Art.  78,  that  the  at- 
traction will  be  directly  as  the  distance  from 
the  centre. 

2.  10  -r-  22  =  2*  Ibs. 

3.  Nothing. 

4.  At  a  distance  from  the  outside  somewhat  less 

than  half  the  thickness  of  the  shell.  The  ex- 
act distance  cannot  be  found  unless  the  thick- 
ness of  the  shell  be  given.  (If  R  be  the  out- 
side radius,  rv  the  inside,  and  r  the  required 
distance  from  the  centre  ;  then 


10 


r2  I? 

or 


which  is  a  cubic  equation,  and  may  be  solved 
by  Cardan's  method.) 

PAGE  44. 

5.  Yes  for  the  first  answer  ;  No  for  the  second. 

6.  On  the  opposite  side.     He  could  not  stop  at  the 


TO  ELEMENTARY  MECHANICS.  49 

centre  by  a  mere  effort  of  the  will.  Some  entertain 
the  idea  that  the  will  of  a  person,  causing  a  movement 
of  parts  of  the  body,  could,  in  a  measure,  control  the 
movement  of  the  body  as  a  whole ;  but,  as  a  fact,  the 
matter  of  the  body  is  subject  to  the  action  of  forces, 
like  any  other  matter.  No  part  of  the  body  can  be 
moved  except  in  accordance  with  the  three  laws  of 
motion.  If  an  arm  is  moved  in  one  direction,  some 
other  part  of  the  body  will  be  moved  in  an  opposite 
direction  (the  body  being  free).  While  passing  across 
the  hollow  referred  to  in  the  Exercise,  the  person 
might  throw  his  arms  about,  or  kick,  or  perform  the 
evolutions  of  swimming,  or  rowing  ;  but  as  there  is 
supposed  to  be  no  matter,  or  no  body,  for  him  to  act 
against,  he  could  not  change  his  rate  nor  direction  of 
movement.  In  drawing  his  feet  forward  he  would 
necessarily  pull  some  part  of  the  body  backward.  It 
is  shown,  by  higher  analysis,  that  the  centre  of  gravity 
of  a  moving  system  is  unaffected  by  the  mutual  ac- 
tions of  internal  forces — so,  in  this  case,  the  motion 
of  the  centre  of  gravity  of  the  person  would  be  en- 
tirely unaffected  by  any  contortions  the  person  might 
make. 

7.  At  the  centre  of  the  sphere.     At  the  same  point. 
Uniform. 

PAGE  43. 

8.  The  ball.     If  the  ball  were  very  small  compared 
with  the  person,  the  movement  of  the  person  might 
be  neglected  compared  with  that  of  the  ball ;  just  as 
the  motion  of  the  sun  is  neglected   compared  with 
that  of  the  planets.     In  the  case  of  the  planets,  the 
pulling  force  is  dependent  upon  their  masses  and  dis- 

3 


50  KEY  AND  SUPPLEMENT 

tances,  but  in  the  case  of  the  person  this  force  is  de- 
pendent upon  his  muscular  exertion. 

9.  He  could  not.  In  the  effort  to  throw  the  ball 
away,  a  force  is  developed  between  the  hand  and  ball, 
which  acts  equally  between  the  ball  and  hand,  but  in 
opposite  directions  in  accordance  with  the  third  law  ; 
and  hence  the  ball  would  move  one  way  and  the  per- 
son the  opposite  way ;  and  both  would  move  in 
straight  lines  in  accordance  with  the  first  law,  and 
their  relative  velocities  will  be  inversely  as  their 
masses  in  accordance  with  the  second  law. 

If  the  person  were  placed  at  rest  in  any  position  in 
the  hollow  and  unable  to  reach  anything,  he  could  not 
turn  over,  nor  change  ends ;  that  is,  if  his  head  were 
towards  the  north  and  his  feet  towards  the  south,  he 
could  not  so  change  as  to  have  his  head  towards  the 

o 

south  and  his  feet  towards  the  north.  Should  he  at- 
tempt to  so  turn  as  to  bring  his  head  towards  the 
south,  he  would  cause  his  f  jet  to  approach  his  head, 
and  they  would  meet  about  half  way.  He  might  suc- 
ceed in  kicking  his  own  head,  or,  if  he  were  very 
flexible,  the  head  and  feet  might  pass  each  other ; 
but  the  body  could  not  turn  over  so  as  to  change  ends. 
Neitlier  could  he  roll  over.  If  one  end  of  the  body 
turns  one  way,  the  other  end  necessarily  turns  the 
opposite  way. 

If  a  cat  be  held  with  her  back  clown  two  feet  or 
so  from  the  floor,  and  dropped,  she  will  strike  on 
her  feet:  how  does  ehe  do  it?  According  to  the 

7  O 

principles  of  mechanics,  if  there  were  nothing  for 
her  to  act  against  it  would  be  impossible  for  her 
to  turn,  and  she  would  necessarily  strike  on  her  back. 


TO  ELEMENTARY  MECHANICS.  51 

While  experimenting,  I  was  surprised  to  find  how 
near  the  floor  the  cat  might  be  held,  and  often 
apparently  perfectly  unwatchful  when  dropped, 
and  yet  alight  on  her  feet.  Now  the  movement  of  the 
body  should  be  accounted  for  on  mechanical  principles. 
Instinct  operates  quicker  than  reason,  and  it  appears 
to  be  certain  that  the  cat,  instinctively,  initiates 
a  rotation  of  her  body  at  the  instant  she  is  dropped. 
"While  it  is  difficult  to  see  how  muscular  action  can 
be  quick  enough  to  produce  this  result,  yet  I  see  no 
other  way  of  accounting  for  the  rotation.  Suspend 
the  cat  with  a  string  at  each  foot,  then  suddenly  cut 
the  strings,  and  she  will  rotate  herself.  It  is  also  an 
interesting  fact  that  she  will  strike  on  her  feet  if  let 
fall  several  feet,  say  six  or  eight  feet.  Now  if  she  had 
the  same  initial  rotation  when  falling  six  feet,  as 
when  falling  two  feet,  why  would  she  not  turn  too 
far  in  the  former  case  and  strike  on  her  side  ?  To  ex- 
plain this,  we  here  state  a  principle  not  yet  proved  in 
this  course.  If  a  rotating  Ijody  self -contracts,  it  will 
rotate  more  rapidly r,  l>ut  if  it  self-expands  it  will  ro- 
tate more  slowly.  The  cat  has  the  ability,  within  a 
limited  range,  of  expanding  or  contracting  the  trans- 
verse dimensions  of  her  -body,  and  to  that  extent  of 
regulating  the  amount  of  her  rotation. 

Consider  still  further  the  relations  of  the  man  and 
ball  in  the  ninth  exercise.  He  puts  his  hand  in  close 
contact  with  the  ball,  but  without  grasping  it,  and 
they  move  away  from  each  other,  until  both  strike 
the  walls  of  the  hollow.  Then  suppose  that  the  man 
springs  for  the  ball  and  seizes  it,  and  then  springs 
towards  the  centre  of  the  sphere,  but  before  reaching 


52  KEY  AND  SUPPLEMENT 

the  opposite  side  throws  the  ball  in  anger.  If  the 
ball  goes  in  a  direction  perpendicular  to  the  line  of 
his  body  (or,  generally,  in  any  line  not  passing 
through  the  centre  of  his  body),  the  man  will  be 
thrown  into  a  rotary  motion  as  well  as  a  motion  of 
translation,  and  he  will  inevitably  perform  somer- 
saults while  backing  away  to  the  opposite  side  of  the 
hollow. 

PAGE  44. 

10.  100  -4-  32£  =  3TVj  =  3-109. 

-..,     n       ..  Mass         200  x  6 

11.  Density  =  •=-.  --  —  =          -^  =  3-109. 

Volume       193  x  2 

10     -,f  2-2  B>  x  5  x  (5 

12.  Mass  =  -  -  =  -342. 


=  -00581' 


14.  If  the  resistance  of  the  air  be  considered  it 
would.  In  the  second  case,  it  would  not  stop, 
but  would  go  from  surface  to  surface  with  the 
regularity  of  a  pendulum.  (See  text,  p.  249.) 
In  the  third  case  the  velocity  would  be  greatest 
at  the  centre,  if  the  hole  be  a  vacuum  ;  but  if 
it  be  filled  with  air,  the  greatest  velocity 
would  be  passed  before  the  ball  reached  the 
centre. 

AKT.  86.  The  value  of  F=  Mf  is  sometimes  called 
the  absolute  measure  of  force,  but  nothing  is 
gained  by  the  term,  except  that  it  distinguishes  it 


TO  ELEMENTARY  MECHANICS.  53 

from  the  mere  stress  which  it  would  produce  if  no 
motion  resulted.  Some  English  writers  call  the 
value  of  F  when  thus  expressed  The  Poundal,  but 
this  term  has  not  come  largely  into  use. 

Observing  that  acceleration  is  the  rate  of  increase 
of  velocity  (Key,  p.  6),  it  follows  that  this  value  of1 
F  is  the  same  as  that  given  by  Newton's  second  law, 
that  the  force  is  proportional  to  the  change  of  mo- 
mentum produced.  But  we  are  confident  that  the 
constant  use  of  the  term  acceleration  for  rate  of 
change  of  velocity  possesses  great  advantage,  since 
rate  of  change  is  liable  to  be  considered  the  same 
as  actual  velocity.  Indeed  some  text  books  assert 
that  the  momentum  of  a  body  is  a  measure  of  the 
force  acting  upon  it — and  call  it  the  second  law. 
Now  a  body  may  have  momentum  when  no  force 
is  acting  upon  it.  It  is  the  rate  of  change  of  mo- 
mentum that  measures  the  force  producing  the 
change ;  in  other  and  better  words,  the  mass  into 
the  acceleration. 

Observe  that  the  establishment  of  this  equation 
contains  a  very  important  principle.  There  is, 
strictly  speaking,  no  relation  between  pounds  and 
feet ;  but  the  ratio  of  two  weights  may  be  the  same 
as  the  ratio  of  two  linear  measures.  A  ratio  is  an 
abstract  number,  and  often  serves  to  connect  con- 
crete quantities,  forming  an  equation.  Thus,  in 
this  case, 

F  f 

-™.  =  a  ratio,  a  mere  number,  =  — . 

g 

The  equation  being  established,  it  is  operated  upon 


54  KEY  AND  SUPPLEMENT 

algebraically.     This  use  of  ratio  has  many  appli- 
cations in  physics. 

ANSWERS  TO  EXAMPLES. 
PAGE  47. 


1.    Velocity  =  y 


2  x  2000  x  193  x  1> 


6  x  100 
1286-6G  =  35-87  feet. 

25   x  193   x   100      965 


2 

80-42  or  SO      'feet. 


o     TT  7     v         //  C*  x  (25-10)x  193x100^ 
3.    Velocity  =  \/   (-  —r,  —  ^7^- 

\  Ox  500  / 


%  193  =  13-88  feet. 

„  100  x  100       Q1  0.  _ 

4.  Force  =  ~-^  —  --^-  =  dl-2o  IDS. 
32  xlO 


SOLUTIONS    OF    PROBLEMS. 

PAGE  50. 

4.  The  tension  equals  the  weight,  P,  plus  the  force 

W  —  P 
•which  will  produce  the  acceleration.    .„       p  g  is  the 

acceleration  when  P  is  raised  vertically.     The  mass 
multiplied  by  the  acceleration  is  the  moving  force,  or 

P     W  —  P  W  —  P 

f-.  •  —  -  ^  g  :  hence  the  tension  is  P  +   ^f      v  P 
g        W  +  P  '  '  +  P 

=   ~*       _  .     Similarly,  it  equals  W  minus  the  accel- 


TO  ELEMENTARY  MECHANICS. 

FP  2TFP 

-*       TTT     _     a    rr  -*• 

erating  force,  or  I       •   ^+  p   '  W+P 


5.  The  effective  moving  force  is  W  —  T,  hence  from 

W 

Prob.2,  W-  T  =  --/. 
ff 

Substitute/=  fa  and  W-T=iW; 
.:  T-^W. 

If  ascending,  T-W  =  —  /,  or  T  -  W  =  I  TF; 

i? 


T=  W  +  the  force  which  will  produce  the  ac- 

W 

celeration  =  W  +  —  \g  =  £ -W. 

9 


EXAMPLES 
PAGE  50. 


2.  ,  =     .^  =  ixlti  x  25  =134^  ft 


-g,2    or  P  - 
-  *'P+   Wg  '  ~ 

I      x  25  +  2  x  10        =  5 


-i|^  x  25  -  2  x  10 
TF—  P 

2°  5  =  *  ^^'  or  P  =  47'57  fts* 


P-  W  2.<P  +  F) 


56  KEY  AND  SUPPLEMENT 

2  x  6-8  x  9    =  Mi6  ft_ 

.       p_,olb 
' 


i  x  4 

5. 

2  x  10  x  P 

P+10 

2  x  10  x  P 

ft  — 

2  x  10  x  P 

p  +  IQ    ~  '    •'•   P     =  10 

2  x  10  x  P  . 

P  +  10      ;'-P=   °° 


If  the  tension  exceeds  20  Ibs.  P  will  be  nega- 
tive. 


.,       ,/2*(P+  IF) 


2  x  10  x  22       0  «-  K  , 

-  =  2-615  seconds. 


ANSWERS  TO  EXERCISES. 

PAGE  51. 

1.  The  balance  will  indicate  no  tension.  This 
question  was  given,  because  the  author  sometimes 
found  that,  in  a  case  like  Fig.  21,  some  students  would 
assert  that  the  tension  of  the  string  ought  to  be  P  / 
but  by  taking  an  extreme  case,  like  the  one  in  the 
exercise,  the  fallacy  would  be  apparent.  One  of  the 
best  conditions  of  mind  for  searching  after  truth  is  to 
be  convinced  of  one's  error.  When  one  admits  that 


TO  ELEMENTARY  MECHANICS.  57 

his  position  is  erroneous,  he  is  generally  in  a  condi- 
tion to  admit  the  truth. 

2.  If  the  acceleration  is  increasing,  the  tension  will 
exceed   the   weight ;    if   uniform   it   will   equal   the 
weight ;  if  decreasing  it  will  be  less  than  the  weight. 

3.  It  will  be  less  than  his  weight  while  descend- 
ing, and  greater  if  ascending.     In  both  cases,  if  the 
motion  be  uniform,  it  will  equal  his  weight. 

4.  The  tension  will  be  less  than  his  weight. 
PAGE  52. 

ART.  91. — It  has  already  been  stated  that,  in  the 
relations  between  force  and  motion,  we  have  four 
fundamental  elements,  force,  space,  time,  and  mass. 
Force  and  matter  are  so  intimately  connected  that  it 
is  impossible  to  completely  divorce  them  ;  but  force 
may  be  abstracted  from  space  and  time — not  in  the 
sense  that  it  can  exist  without  them,  but  in  the  sense 
that  it  may  be  considered  independently  of  them,  and 
when  so  considered  it  is  called  stress.  But  a  force 
acting  upon  a  body  may  move  it  through  space,  and 
the  space  may  be  considered  independently  of  time. 
The  product  of  force  and  space  both  considered  in  the 
same  line,  and  abstracted  from  time,  is  work. 

The  term  originated  from  the  grosser  ideas  of  labor, 
but  the  definition  given  in  the  text  is  applicable  to 
the  most  refined  actions  in  mechanics.  All  known 
forces  in  nature  are  constantly  working.  Thus,  riv- 
ers wear  the  beds  of  their  streams;  wind  drives  the 
sail,  uproots  trees,  produces  drifts  of  sand,  etc. ;  heat 
expands  bodies,  and  may  overcome  their  cohesion, 
etc. 

According   to   the   definition,  a  man  merely  sup- 
3* 


58  KEY  AND  SUPPLEMENT 

porting  a  weight  cannot  be  said  to  work,  and  yet  he 
soon  becomes  conscious  of  fatigue.  But  a  more  criti- 
cal examination  of  his  case  shows  that  the  weight  is 
not  strictly  at  rest.  The  natural  action  of  his  organ- 
ism, especially  the  beating  of  his  heart,  causes  slight 
elevations  and  depressions  of  his  load,  so  that  he  is, 
in  the  strictest  sense,  constantly  laboring. 

The  following  are  some  examples,  of  the  average 
work  accomplished  by  a  man  under  various  condi- 
tions. 

WORK  OP  MAN  AGAINST  KNOWN  RESISTANCES. 

hrs.  per         ft.  Ibs. 
Ibs.     day.          per  day. 

1.  Raising-  his  own  weight  up  sfcairs  or 

ladder ' 145      8      2,088,000 

2.  Hauling  up  weights  with  rope,  and 

lowering  the  rope  unloaded 40      6         648,000 

3.  Lifting  weights  by  hand 44      6         522,720 

4.  Carrying  weights  up  stairs  returning 

unloaded 143      6         399,600 

5.  Shoveling  up  earth   to  a  height  of 

5ft.  3  in 6    10         280,800 

6.  Wheeling  earth  in  barrow  up  slope 

of  1  in  12,  one-half  horiz.  veloc.  0.9 

ft.  per  sec. ,  and  returning  unloaded,  132    10         356,400 

7.  Pushing  or  pulling  horizontally  (cap- 

stan or  oar) 26.5    8      1,526,400 

8.  Turning  a  crank  or  winch 18       8      1,296,000 

9.  Working  pump 13.2  10      1,183,000 

10.  Hammering 15       8?       480,000 

PERFORMANCE  OF  A  MAN  TRANSPORTING  LOADS  HORIZONTALLY. 

hrs.  per      Ibs.  con'd 
Ibs.     day.  1  ft. 

11.  Walking  unloaded,  transport  of  own 

weight 140     10    25,200,000 


TO  ELEMENTARY  MECHANICS.  59 

hrs.  per    Ibs.  con'd 
Ibs.    day.  1  ft. 

12.  Wheeling  load  in  2- wheel  barrows,re- 

turn  unloaded 224    10    13,428,000 

13.  Wheeling  load  in  1-wheel  barrow,  re- 

turn unloaded 132    10      7,920,000 

14  Traveling  with  burden 90      7      5,670,000 

15.  Carrying  burden,  returning  unloaded.  140      6      5,032,800 

See  Rankine's  Steam  Engine,,  pp.  84-85,  where  the 
rate  of  doing  the  above  works  is  also  given. 

Morin  and  Weisbach  give  2,387  ft.  Ibs.  per  minute 
as  the  work  which  a  man  is  capable  of  doing  when 
working  eight  hours  consecutively.  This  equals  -jVoVt 
=  0.07  +  ILP.  nearly. 

At  an  experiment  made  at  Dresden  in  1 880,  men 
working  only  2  minutes  at  a  time  on  a  hand  fire-en- 
gine did  0-277  H.P. — or  nearly  four  times  that  given 
above. 

PAGE  54,  ART.  98. — Nystrom  asserts  in  his  writings, 
that  work  is  not  independent  of  time,  for  it  requires 
time  to  move  a  body  over  space ;  also  that  if  one 
horse  drew  twice  the  load  over  the  same  space  as  an- 
other, he  did  twice  the  work  and  was  twice  as  effi- 
cient. But  we  hold,  and  trust  we  have  clearly  shown, 
that  time  is  properly  abstracted  in  the  idea  of  work— 
that  efficiency  is  very  different  from  work.  We  have 
also  shown  that  if  time  is  considered  even  implicitly, 
the  velocity  must  also  be  considered. 

AET.  99. — The  simple  definition — Power  is  rate  of 
doing  work — is  coming  more  generally  into  use. 

PAGE  56,  ART.  104. — Friction  is  a  force;  its  value  can 
be  measured  in  pounds.  It  does  not  directly  produce 
a  positive  acceleration,  but  a  negative  one.  Its  direct 
office  is  to  destroy  motion  ;  not  to  produce  it.  But 


60  KEY  AND  SUPPLEMENT 

indirectly  it  may  produce  motion  by  produc  ng  heat. 
Heat  produces  motion,  and  the  work  of  friction  lias 
its  equivalent  in  heat,  and  this  heat  if  collected  would 
produce  the  same  motion  as  that  which  it  has  de- 
stroyed. But  in  practice  it  is  so  quickly  dissipated 
that,  in  most  cases,  it  is  apparently  lost. 

A  smooth  surface  is  one  from  which  the  idea  of 
roughness  is  abstracted. 

PAGE  58,  ART.  107. — The  laws  of  Moriu  are  only  ap- 
proximately correct.  In  machinery,  the  character  of 
the  surfaces  in  contact,  the  mechanical  execution  of 
the  fitting  up,  and  of  the  lubricants,  are  each  and  all 
important  elements.  See  practical  treatises  and  arti- 
cles upon  the  subject.  Rankine,  Steam  Engine,  pp. 
14-18,  Thurston,  Friction  and  Lubrication. 

PAGE  59,  ART.  109. — The  frictional  resistance  of  rail- 
road trains  is  principally  rolling  friction  under  good 
working  conditions.  A  railroad  train  in  good  order, 
and  on  a  good  road,  will  not  be  safe  against  starting 
under  the  action  of  gravity  alone,  unless  the  gradient 
is  less  than  eighteen  or  twenty  feet  to  the  mile  ;  once 
started  it  will  continue  in  motion  on  gradients  as  low 
as  thirteen  feet  to  the  mile.  The  coefficient  of  rolling 
friction  for  trains  in  good  order  is  ^^  =  0-0025,  or 
less  than  six  pounds  per  ton.  The  resistance  at  start- 
ing is  -yflj.  =  0-0038  or  8£  pounds  per  ton. 

The  resistance  of  a  locomotive  is  about  12  pounds 
per  ton. 

The  resistances  on  railroads,  under  average  condi- 
tions, and  including  all  forms  of  resistances,  are  given 
by  Clarke. 


TO  ELEMENTARY  MECHANICS.  61 

"When  the  permanent  way  is  straight,  rails  dry  and 
clean,  he  gives  for  trains  only 


for  engine  and  train 


171  ' 


where  R  is  in  pounds  per  ton  gross,  and  v  the  velocity 
of  the  train  in  miles  per  hour.  (Manual  of  Rules,  etc., 
p.  965.)  A  Mr.  Hughes  found  on  an  English  "tram- 
way "  a  resistance  of  twenty-six  pounds  per  ton. 

On  railroads,  frictional  resistances  are  sometimes 
greatly  increased  by  the  resistance  of  the  air,  called 
"  head  resistance,"  and  amounts,  in  pounds  per  square 
foot  of  front  exposed,  to  0-005  of  the  square  of  the 
velocity  in  miles  per  hour  with  which  the  air  meets  the 
head  of  the  train.  Side  winds  often  increase  the 
flange  resistance  seriously. 

The  value  of  the  coefficient  of  friction  on  ordinary 
railroads  is  0-003,  on  well  laid  railroad  tracks  0-002, 
on  best  possible  railroad  0-001. 

Mr.  S.  Whinery  (Trans.  Am.  Soc.  C.  E.,  April, 
1878)  gives  a  formula  for  the  total  resistances  of  a 
train  running  on  curves, 


where  R  =  total  resistance,  D  =  degrees  of  curvature, 
g  =  gauge  of  track,  t  =  length  of  rigid  wheel  base,  a 


62  KEY  AND  SUPPLEMENT 

and  n  are  quantities  expressing  resistances  due  to  ac- 
cidental and  irregular  conditions.  These  resistances  are 
inversely  as  the  radius  of  curvature,  directly  as  the 
load,  and  nearly  independent  of  the  velocity. 

Mr.  O.   Chanute   (Trans.  Am.  Soc.  C.  E.,    April, 
1878)  analyzes  this  increase  of  resistance  as  follows : 


Due  to  twist  of  wheel 0-001 

"      slip        "         0-1713 

"      flange  friction 0-2450 

"      loss  of  couplings 0-0213 


Total..  .  04386 


Loose  wheels  reduce  this  loss  20  or  25  per  cent.  The 
rigid  form  of  wheel-base  of  European  cars  and  loco- 
motives doubles  the  increase  due  to  curves  as  well  as 
increases  the  resistance  on  the  straight  line.  Accord- 
ing to  Mr.  Chanute  the  "  coning  "  of  wheels  increases 
the  resistance  from  0-125  to  0-25  pounds  per  degree 
of  curve  per  ton.  (Thurston,  Friction  and  Lubrica- 
tion, pp.  13-18.) 

Recent  experiments  on  the  New  York  and  Erie 
R.  R.  show  that  on  a  track  of  steel  rails  in  first-class 
condition,  the  friction  of  a  train  at  low  velocities  may 
be  reduced  to  3£  or  4  pounds  per  ton  on  a  horizontal 
road ;  and  that  the  rolling  resistance  on  such  a  track 
in  the  summer  may  be  safely  taken  at  5  pounds  per 
ton.  (R.  R.  Gazette,  March  24,  1882,  or  Haswell's 
Pocket- Book  for  Engineers,  edition  of  1882.) 


TO  ELEMENTARY  MECHANICS.  63 

PAGE  63. 

Example  3. — If  /<  =  0, the  expression  /jf> W  +  hW be- 
comes li  W,  in  which  case  the  work  equals  that  neces- 
sary to  raise  the  weight  through  the  height  of  the 
plane. 


SOLUTIONS  OF  EXAMPLES. 
PACK  64. 

1.  It  will  raise  50  X  33,000  x  60  ft.  Ibs.  in  1  hour, 

which  divided  by  500    x  62-5  will  give  the 
cubic  ft.  =  3,168. 

2.  The  average   height  to   which  the  material  is 

raised   will  be  10  ft.     Hence  the  work  =  140 
(i  7t  x  32  x  20)  x  10  =  197,920-8.  Ibs. 

F  +  F'         1000  +  200      10      _0. 
3'  8  =  -W  a=  —2000-  X  12  =  7'2m- 

4.  39-37  inches  =  3  2808  ft.  x  2-2  Ibs.   =  7-217 

+  ft.  Ibs. 

5.  43,333  x  7-217  +  ~  (32,808)2  =  29,057  ft.  Ibs. 

6.  Substituting  in  the  answer  to  Prob.  2,  p.  61,  we 

} 

have  H.  P.  =  0-9114  x  1  x  12  =  37-89. 

7.  Work  of  the  fall  =  2,000  x  8  =  16,000  ft.  Ibs. 

Let  x  —  the  distance  driven,  then   10,000a;  = 
16,000;  .-.  #  =  1-6  ft. 

8.  Find  the  velocity  in  feet  per  minute.     We  have 


64:  KEY  AND  SUPPLEMENT 

2  x  5280        rp,      ,  „ 

v  =  -  —  --  .     I  he  horse-power  =  Jrv  -f- 
60 

200  x  2  x  5280 
33'000>°r     60x33000        -1*- 

Q        _  y^P  -2(P  +  TT)s_  1^x9x8-2(8  +  40)4 

^•JF  i£i  x  9  x  40 

=  0-1668. 
PAGE  65. 

P-  HW    „      t       5-25x045 
10.  6-=  i.-^—^tf..  5+25 


x  x  25  =  1206-25  H-  72  =  16-75  ft. 

6 


ANSWERS   TO   EXERCISES. 

1.  One  pound  raised  one  foot.     Work   is  a  com- 

pound quantity,  compounded  of  stress  and 
space. 

2.  See  preceding  remarks,  text,  p.  56. 

3.  It  is  dependent  upon  time  only  implicitly,  and 

in  the  sense  that  motion  requires  time.  But, 
strictly,  time  should  be  abstracted. 

4.  Tang.  15°  =  0-268. 

5.  5  Ibs. 

6.  Mechanical  power  is  not  work,  but  rate  of  doing 

work — just  as  velocity  is  not  space,  but  rate  at 
which  space  may  be  passed.  Mechanical  power 
involves  a  unit  of  time,  but  work  does  not. 

PAGE  66. 

ART.  41. — The  doctrine  of  energy  is  the  grand,  gen- 
eral principle  of  modern  physics.     All  the  changing 


TO  ELEMENTARY  MECHANICS.  65 

phenomena  of  nature  are  but  manifestations  of  the 
transmutation  of  energy.  Its  principles  are  not  de- 
duced by  any  system  of  mathematics,  but  by  a  long 
series  of  inductions.  We  accept  its  general  principles 
without  attempting  a  general  demonstration. 

Even  work  in  a  higher  sense  is  but  a  means  of 
transmitting  energy.  Thus,  a  horse  works  by  draw- 
ing a  load,  but  it  is  simply  a  means  of  transmitting 
the  energy  possessed  by  the  horse,  first  into  energy 
stored  in  the  mass  of  the  load,  and  second  into  heat 
by  means  of  the  friction  overcome.  Still,  work  is  not 
only  a  convenient,  but  a  useful  term.  Work  done  is 
one  idea,  energy  produced  as  its  equivalent  is  another 
—force  and  space  are  the  elements  of  the  former, 
mass  and  velocity  of  the  latter. 

DEFINITIONS    OF    WORK    AND    ENERGY. 

"  Work  is  the  overcoming  of  resistance  continually 
recurring  along  some  path." — BARTLETT'S  Elements 
of  Analytical  Mechanics,  p.  26. 

"  Work  consists  in  moving  against  resistance.  The 
work  is  said  to  be  performed,  and  the  resistance 
overcome." — RANKINE'S  Applied  Mechanics,  p.  477. 

"  Work  is  the  effect  of  strain  and  motion  combined. 
—  -TRAUTWINE'S  Engineer^  Pocket-Book,  p.  445. 

Remarks  on  above  from  p.  446  same  book  :  "  Grav- 
ity acting  in  a  body  falling  freely  in  a  vacuum,  and 
consequently  unresisted,  exerts  no  effort  upon  it,  it 
neither  goes  before,  and  pulls  it  along,  nor  behind, 
and  pushes  it ;  for  there  can  be  no  pull  or  push  except 


66  KEY  AND  SUPPLEMENT 

when  there  is  some  force  to  pull  or  push  against. 
But  it  simply,  as  it  were,  animates  the  body,  or  en- 
dows it  with  the  power  of  locomotion.  As  the  body 
falls,  the  force  of  gravity  which  gives  it  motion  all 
remains  unimpaired,  and  stored  up  in  it,  ready  to  ex- 
ert an  effort  against  any  other  force  which  it  may 
chance  to  meet  with.  Therefore  a  body  falling  unre- 
sistedly  has  no  weight ;  for  gravity,  which  gives  it 
weight  alone  while  at  rest,  now  gives  it  motion 
alone." 

(The  word  strain  in  the  above  definition  is  improp- 
erly used  for  stress. — AUTHOR.) 

"  A  force  is  said  to  do  work  if  its  place  of  applica- 
tion has  a  positive  component  motion  in  its  direction ; 
and  the  work  done  by  it  is  measured  by  the  product 
of  its  amount  into  this  component  motion." — THOMSON 
AND  TAIT,  Nat.  Philos.,  p.  176. 

"  Work  done  on  a  body  by  a  force  is  always  shown 
by  a  corresponding  increase  of  vis  viva,  or  kinetic  en- 
ergy, if  no  other  forces  act  on  the  body  which  can  do 
work  or  have  work  done  against  them.  If  work  be 
done  against  any  force,  the  increase  of  kinetic  energy 
is  less  than  in  the  former  case  by  the  amount  of  work 
so  done.  In  virtue  of  this,  however,  the  body  pos- 
sesses an  equivalent  in  the  form  of  potential  energy, 
if  its  physical  conditions  are  such  that  these  forces 
will  act  equally,  and  in  the  same  directions,  if  the 
motion  of  the  system  is  reversed." — /&.,  p.  177. 

"  An  agent  is  said  to  do  work  when  it  causes  the 
point  of  application  of  the  force  it  exerts  to  move 
through  a  certain  space.  Motion  is  essential  to 


TO  ELEMENTARY  MECHANICS.  67 

work." — TWISDEN,  Elementary  Introduction  to  Prac- 
tical Mechanics,  p.  18. 

"  Mechanical  effect,  or  work  done,  is  that  effect 
which  a  force  accomplishes  in  overcoming  a  resist- 
ance. It  depends  not  only  on  the  force,  but  also  on 
the  space  during  which  it  is  in  action,  or  during 
which  it  overcomes  a  resistance." — WEISBACH'S  Me- 
chanics of  Engineering^  p.  168. 

"  Work  is  the  production  of  motion  against  resist- 
ance."— TODHUNTER,  Mechanics  for  Beginners,  p.  337. 

"Work,  same  as  before.  According  to  this  defini- 
tion, a  man  who  merely  supports  a  load  does  not 
work ;  for  here  there  is  resistance  without  motion. 
Also  while  a  free  body  moves  uniformly  no  work  is 
performed ;  for  here  there  is  motion  without  resist- 
ance."— TODHUNTEK,  Nat.  Philos.  for  Beginners. 
p.  255. 

"  Whenever  a  body  moves  through  any  space  in  a 
direction  opposite  to  that  in  which  a  force  is  acting 
on  it,  work  is  said  to  be  performed.  It  is  evident 
that  the  application  of  force  is  necessary  to  overcome 
resistance,  and  it  is  very  often  found  convenient  to 
measure  the  work  done  by  the  amount  of  force  ex- 
pended, and  the  distance  in  the  direction  of  the  force 
through  which  it  has  been  employed." — MAGNUS'S 
Elementary  Mechanics,  p.  102. 

"  Thus  the  increase  of  vis  viva,  which  is  also  the 
work  done  by  the  acting  forces  on  the  body." — 
PRICE'S  Infinitesimal  Calculus,  vol.  iii.,  p.  636. 

"  Work  is  done  when  resistance  is  overcome,  and 


68  KEY  AND  SUPPLEMENT 

the  quantity  of  work  done  is  measured  by  the  prod- 
uct of  the  resisting  force  and  the  distance  through 
which  that  force  is  overcome." — MAXWELL'S  Theory 
of  Heat,  p.  87. 

"  Work  is  the  overcoming  of  mechanical  resistance 
of  any  kind." — NYSTKOM'S  Pamphlet  on  Force  of 
Falling  Bodies,  etc.,  p.  25. 

DEFINITIONS   OF    ENERGY   AND  VIS   VIVA. 

"  Energy  expresses  power  to  do  work,  or  force  stored 
and  ready  for  use." — McCuLLocn's  Treatise  on  Me- 
chanical Theory  of  Heat,  p.  40. 

"  Vis  viva  (energy)  is  a  quantity  which  varies  as 
the  product  of  the  mass  of  a  particle  and  the  square 
of  its  velocity." — PRICE'S  Infinitesimal  Calculus,  vol.  3, 
p.  360.— 2d  Ed.  Oxford. 

"  Living  force,  or  vis  viva  (or  energy),  is  nothing 
more  than  an  expression  referring  to  the  quantity  of 
work  (motion  and  strain  combined)  which  the  force 
in  a  body  at  any  given  instant  could  perform,  if  left 
to  itself,  without  afterwards  receiving  any  additional 
force." — TRAUTWINE,  Civil  Eng.  Pocket  Book,  p.  446, 
Ed.  1872. 

"Energy  measures  the  quantity  of  working  power 
of  a  moving  body." — BARTLETT,  Elements  of  Anahjt. 
Mech.,  9th  Ed.,  p.  116. 

"  The  product  of  the  mass  of  a  body  by  the  square 
of  its  velocity  is  called  its  living  force  or  vis  viva" — 
Jlecanique  Celeste,  p.  99. 


TO  ELEMENTARY  MECHANICS.  69 

"  Energy  means   capacity  for  performing  work." — 
RANKINE,  Applied  Mechanics,  p.  477. 

"  Vis  viva,  or  living  force  (or  energy),  is  the  power 
of  a  moving  body  to  overcome  resistance,  or  the 
measure  of  work  which  can  be  performed  before  the 
body  is  brought  to  a  state  of  rest." — SILLIMAN,  Prin- 
ciples of  Physics,  2d  Ed.,  p.  78. 

"Energy  is  the  capacity  a  body  has,  when  in  a 
given  condition,  for  performing  a  certain  measurable 
quantity  of  work." — TODHUNTER,  Natural  Philosophy 
for  Beginners,  p.  264. 

"Energy  of  a  body  is  power  of   doing  work."- 
MAGNUS,  Lessons  in  Elementary  Mechanics,  p.  110. 

"  Energy  of  a  body  is  the  capacity  which  it  has  of 
doing  work,  and  is  measured  by  the  quantity  of  work 
which  it  can  do.  The  kinetic  energy  of  a  body  is  the 
energy  which  it  has  in  virtue  of  being  in  motion." — 
GUMMING,  Theory  of  Electricity,  p.  5. 

"  Kinetic  energy  or  vis  viva  is  defined  as  half  the 
product  of  the  mass  into  the  square  of  the  velocity." — 
Ibd.,  p.  13. 

"  Energy  is  defined  to  be  capacity  for  doing  work." 
"  It  is  of  two  kinds — kinetic  or  actual  when  the 
body  is  in  actual  motion.  Potential  or  latent  when 
the  body,  in  virtue  of  work  done  upon  it  occupies  a 
position  of  advantage,  so  that  the  work  can  be  at  any 
time  recovered  by  the  return  of  the  body  to  its  old 
position." — Ibd.,  p.  14. 

"Energy   is   the   capacity  of  a  body  to   perform 


70  KEY  AND  SUPPLEMENT 

work.  Energy  is  said  to  be  stored  when  this  capacity 
is  increased,  and  to  be  restored  when  it  is  diminished. 
The  units  of  work  and  of  energy  are  the  same."  — 
WEISBACII,  Mech.  and  Eng.  Translator's  note,  bottom 
of  page  168. 

"  If  we  adopt  the  same  units  of  mass  and  velocity 
as  before  there  is  particular  advantage  in  defining 
kinetic  energy  as  half  the  product  of  the  mass  into 
the  square  of  the  velocity."  —  ROUTE'S  Rigid  Dynam- 
ics. 

PAGE  67. 

AET.  112.  —  If  the  body  had  an  initial  velocity  v0, 
the  work  done  upon  it  in  passing  from  a  velocity 
V0  to  V,  or  the  work  which  would  be  given  out  by  it 
in  passing  from  the  velocity  v  to  v0,  will  be 


This  may  be  deduced  in  another  way,  if  we  antici- 
pate the  equations  for  momentum. 
Thus,  Eq.  (2),  p.  78,  of  the  text  is 

Ft  =  M(v  -  v0). 

The  space  over  which  F  acts  will  be  equivalent  to 
the  average  velocity  into  the  time,  or 

s  =  \(v  +  vjt. 

Multiplying  these   equations,   member  by   member, 
and  canceling  £,  gives 

Fs  = 


TO  ELEMENTARY  MECHANICS.  71 

as  before.  But  we  do  not  consider  this  method  as 
good  as  the  first,  for  when  the  velocity  changes  irreg- 
ularly, it  is  not  so  evident  that  the  space  equals  the 
average  of  the  extreme  velocities  into  the  time. 

O 

ART.  113. — Potential  energy  is  relative.  Thus  if  a 
body  whose  weight  is  10  Ibs.  is  40  feet  above  the 
earth,  its  potential  energy  in  reference  to  the  earth 
is  10  x  40  =  400  ft.  Ibs. ;  but  if  it  be  over  a  well  20 
feet  deep,  the  potential  energy  in  reference  to  the 
bottom  of  the  well,  will  be  10  x  60  =  600  ft.  Ibs. ; 
and  in  reference  to  its  own  position  it  is  nothing.  It 
is  the  work  which  the  body  may  do  in  reference  to 
some  point — or  condition — arbitrarily  chosen. 

PAGE  69. 

ART.  115. — For  the  mathematical  theory  of  heat, 
see  Poisson  Traite  des  Cheleur,  Fourier  Theorie  de 
la  Cheleur,  Rankine  on  The  Steam  Engine,  Clausius 
on  The  Theory  of  Heat,  Hirn's  investigations,  Max- 
well's Theory  of  Heat,  McCulloch's  Theory  of  Heat, 
Tait's  History  of  Thermodynamics,  etc.,  etc. 

PAGE  72. 

ART.  117. — This  is  one  of  the  most  important  phys- 
ical constants  which  has  been  determined  in  recent 
times.  For  a  comprehensive  and  able  review  of  the 
methods  by  which  it  has  been  determined,  as 
well  as  for  a  description  of  that  author's  methods 
and  results,  see  Mechanical  Equivalent  of  Heat,  by 
Professor  II.  A.  Eowland.  (Proceedings  of  the 
American  Academy  of  Arts  and  Sciences,  1879,  p. 
45.)  Prof.  Rowland's  results  differ  by  only  a  small 
per  cent,  from  those  given  by  Joule.  He  states, 


72  KEY  AND  SUPPLEMENT 

p.  44,  that  the  value  found  by  Joule  at  14°,  agrees 
with  his  results  at  18°  C.     The  value  772  foot  pounds 
still  stands  as  a  practically  correct  one. 
PAGE  73. 

ART.  118. — In  regard  to  energy  generally,  it  appears 
that  all  of  it — or  at  least  very  nearly  all — originates 
with  the  sun.  It  was  a  beautiful  remark  of  John 
Stephenson — as  he  saw  a  railroad  train  winding  its 
way  through  the  country — "  that  train  is  drawn  by 
the  heat  of  the  sun."  The  heat  and  light  of  the  sun 
caused  the  growth  of  vegetation  ;  that  vegetation  in 
time  was  gathered  into  great  masses,  which  in  time 
became  coal  in  the  mine.  This  coal  was  brought 
forth  and  used  as  fuel  in  the  locomotive ;  so  that  it 
originated  in  the  action  of  the  sun.  All  energy  on 
the  earth  is  due  to  the  light  and  heat  of  the  sun. 
Activity  in  the  commercial  world  is  directly  depend- 
ent upon  it ;  for  if  the  sun,  on  account  of  spots  upon 
it,  or  from  other  causes,  does  not  dispense  with  its 
usual  heat,  short  crops  will  result,  and  thus  affect  all 
the  business  of  a  country,  if  not  of  the  world.  In  this 
way  the  sun  may  be  charged  with  causing — more  or 
less  directly — depressions  in  trade,  or  activity  in  com- 
merce, as  the  case  may  be ;  and  hence,  in  some  cases 
at  least,  of  producing  sadness  or  cheerfulness  in  the 
home  circle.  Even  the  religious  world  is  affected  by 
the  action  of  the  sun.  In  earlier  times,  on  account 
of  the  superstitions  of  the  people,  religious  leaders 
appealed  more  or  less  effectually  to  the  fears  of  their 
followers  when  the  sun's  rays  were  cut  off  by  an 
eclipse ;  and  in  modern  times  the  feeling  of  depend- 
ence upon  the  Creator  is  too  often  modified  by  the 


TO  ELEMENTARY  MECHANICS.  73 

prosperity  or  depression  caused  by  the  circumstances 
surrounding  them,  the  conditions  of  which  were  de- 
pendent upon  the  elements  of  nature,  and  these 
caused,  more  or  less  directly,  by  the  action  of  the  sun. 
The  sun  appears  to  be  dispensing  its  energy  to  his 
family  of  planets,  and  in  this  way  wasting  itself 
away.  Sir  Isaac  Newton  realized  this  condition  of 
things  and  saw — or  thought  he  saw — the  necessity 
of  the  sun's  being  replenished  in  order  to  maintain  its 
stock  of  energy ;  and  he  conceived  that  this  might  be 
done  by  one  comet  after  another  falling  into  the  sun. 
As  the  comets  come  from  remote  regions  of  space, 
they  would  possess  a  large  amount  of  energy  when 
they  struck  the  sun,  and  by  falling  into  it  would  pro- 
duce intense  heat.  No  comet,  however,  has  within 
historical  times  been  known  to  fall  into  the  sun  ;  but, 
on  the  other  hand,  the  most  critical  examination  of 
their  orbits  shows  that  their  paths  are  nearly  as  well 
defined — and  nearly  as  fixed  in  position — as  any  of 
the  planets  ;  and  no  cause  is  known  to  exist  that  will 
cause  them  to  fall  into  the  sun.  It  was  formerly  sup- 
posed that  the  ether  of  space  caused  a  resistance  to 
the  motion  of  all  bodies  in  space;  and  if  it  did,  not 
only  would  the  comets,  but  also  the  planets,  ultimately 
fall  into  the  sun.  However,  nothing — absolutely  noth- 
ing— is  known  in  regard  to  the  effect  of  this  ether 
upon  the  motion  of  bodies  in  it.  Comets  and  planets 
have  moved  in  their  orbits  for  untold  ages,  and,  for 
aught  we  know,  have  maintained  their  relative  posi- 
tions. If  the  comets  were  destined  to  fall  into  the 
sun,  they  must  have  been  doing  so  for  ages  and  ages, 
and  hence  must  originally  have  been  comparatively 
4 


74  KEY  AND  SUPPLEMENT 

very  numerous.  It  would  seem  that  sufficient  time 
had  elapsed  since  the  existence  of  the  solar  system 
to  have  exhausted  this  stock  of  energy — still,  at  the 
present  time,  comets  are  numerous.  Similarly,  in  re- 
gard to  the  planets,  the  most  refined  observations, 
combined  with  the  most  refined  analysis,  have  failed  to 
detect  any  modification  of  motion  due  to  the  ether 
of  space. 

More  recently,  the  late  Professor  Benj.  Pierce,  of 
Cambridge,  Mass.,  put  forth  the  theory  that  the  en- 
ergy of  the  sun  was  supplied  by  meteorolites  tailing 
from  an  immense  distance  directly  into  the  sun.  The 
meteors  are  dark  bodies,  and  it  is  assumed  that  there 
may  be  multitudes  of  them  scattered  through  space. 
That  there  are  many,  is  shown  by  the  fact  that  they 
frequently  fall  upon  the  earth.  Admitting  the  truth 
of  this  hypothesis,  it  seems  inevitable  that,  in  the 
course  of  time,  the  supply  will  be  exhausted — and  then 
the  question,  What  will  follow?  becomes  a  serious 
one  to  science. 

Space,  which,  in  our  younger  days,  we  conceived 
to  be  void,  is  really  filled  with  something,  and  there 
may  be  vastly  more  inert  matter  scattered  through 
it  than  we  have  imagined. 

A  mere  contraction  of  the  volume  of  the  sun, 
caused  by  the  mutual  attraction  of  its  own  particles, 
will  produce  heat.  The  author  has  shown  that  if  the 
earth  contracted  from  twice  its  present  diameter  to  its 
present  size,  and  all  the  energy  thus  produced  be 
changed  into  heat,  and  uniformly  disseminated 
throughout  the  mass,  the  temperature  would  be  raised 
44,655  degrees  F.,  if  it  had  the  specific  heat  of  water, 


TO  ELEMENTARY  MECHANICS.  75 

or  357,240  degrees  F.,  if  it  had  the  specific  heat  of 
iron.  (See  Analytical  Mechanics,  p.  229,  or  Mathe- 
matical Visitor,  1880,  p.  134.) 

We  believe  that  the  solar  system  is  stable,  that  it  is 
not  made  to  run  down,  that  it  has  the  elements  of 
self-preservation  ;  but  we  cannot  prove  it.  Neither 
can  those  who  entertain  the  opposite  view  prove  their 
position.  This  problem  is,  at  present,  beyond  the 
reach  of  science. 

SOLUTIONS  OF  EXAMPLES. 
PAGE  75. 

25 

1.  — -  =  the   mass,  and  as  32|  is  the  acceleration 
32% 

in  feet  per  second,  the  velocity  should  be  in 
the  same  units  ;  hence  v  =  W  =  |  feet  per 
second,  and  the  work  will  be 

OK        /S\2 

'•J|-(t)=m  +  f'-lbS- 

2.  The  work  will  vary  as  the  depth,  and  the  energy 

as  the  square  of  the  velocity ;  hence, 

depth  =  2(y)2  =  18  feet. 

PAGE  76. 

3.  Reducing  the  tons  to  pounds,  we  have 

60  tons  =  60  x  2000  Ibs.  =  120,000  Ibs. 

Similarly, 

40  x  5280  ,, 

40  miles  per  hour  =  — — — —  ft.  per  second 

60  x  60 


KEY  AND  SUPPLEMENT 
The  friction  =  60  x  8  =  480  Ibs. 

Let  x  =  the  required  distance  in  miles  =  5280# 
feet,  then  the  work  will  be  480  x  5280  x ; 
hence  we  have 

480  x  5280,  =  i.12«p(*L 

which  solved  gives 

100  /:LPy=  2-53  miles. 


4.  Let  x  =  the  required  number  of  pounds  of  wa- 
ter ;  then  the  energy  put  into  water  in  raising 
it  from  32°  F.  to  212°  will  be 

772  (212  -  32)aj  =  772  x  180*. 
The  kinetic  energy  stored  in  the  train  will  be 

200000  /30  x  5280V. 
32£     \  60  x  60    /  ' 

which,  by  the  conditions  of  the  problem,  equals 
the  heat  energy  to  be  put  into  the  water; 
hence 

200000  /88\2 
772  x  180a;  =  \  x  I  "a"  J  > 

OA-g        \  O  / 

.-.  x  =  19-2  Ibs. 
PAGE  76. 

-    f_P  -  »W     _  4  -  0-2  x  10     193  _  193 
^    =  P  +   W  g  ~  14  6      =  42 

=  4-6; 


TO  ELEMENTARY  MECHANICS.  77 

also  for  the  velocity 

v*  =  2/6-  =  2  x  4-6  x  5  =  46. 
The  work  will  be 

10  x  //  x  distance, 
and  the  energy  will  be 


.•.  0-2  x  10  x  distance  =  $  x  x  46  ; 

j.yo 

/.  distance  =  3^  feet. 

6.  The  friction  =  200  x  0-2  =  40  Ibs.  Let  the 
velocity  per  minute  be  #,  then  the  work  per 
minute  will  be  40#,  and  for  3  minutes  it  will 
be  120a?.  The  energy  of  one  pound  of  water 
raised  one  degree  F.  is  772  ft.  Ibs.,  and  of  5  Ibs. 
it  will  be  5  x  772,  and  for  50  degrees  it  will 
50  x  5  x  772. 
Hence 

120»  =  50  x  5  x  772  ; 
/.  x  =  1,608  ft.  per  minute. 

ANSWERS  TO  EXERCISES. 

PAGE  76. 

1.  It  is  not  ;  force  is  only  one  of  the  elements  pro- 
ducing energy. 

2.  Ability  to  do  work.     "Work  has,  however,  been 
done. 

3.  It   produces    action    of    the   stomach,  thus   in- 


78  KEY  AND  SUPPLEMENT 

volving  energy  ;  promotes  action  of  the  heart ;  causes 
the  growth  of  the  bone,  muscle,  and  flesh,  and  these 
enable  the  animal  to  move  about — to  do  work — to 
swallow  more  food — to  lie  down — to  get  up,  etc.,  etc. 

4.  Because,  in  the  first  place  it  is  not  as  concentra- 
ted, and,  in  the  second  place,  the  heat  is  more  quickly 
conducted  away. 

5.  It  will.     It  is  due  to  this  cause  that  meteors  are 
visible.     The  meteorolites  which  fall  upon  the  earth 
have  the  appearance  of  having  been  partly  melted, 
and  hence  must  have  been   subjected  to  great  heat. 
This  is  due  to  the  compression  of  the  air  in  front  of 
the  meteor  and  of  the  friction  of  the  air  against  its 
sides  as  it  passes  swiftly  through  the  air,  and  the  heat 
thus  produced  is  so  great  that  the  meteor  is  heated  to 
redness,  and  thus  appears  like  a  shooting  star,  as  it 
really  is.     It  is  probable  that  the  smaller  meteorolites 
become  so  nearly  consumed  by  the  great  heat  that 
they  could  scarcely  be  found  after  they  had  fallen 
upon  the  earth,  but   larger   ones   have   been   found 
which  struck  the  earth  with  such  violence  that  they 
nearly  or  quite  buried  themselves.     It  has  been  sug- 
gested that  the  great  iron  deposit  in  the  upper  pen- 
insula of  Michigan,  a   short    distance   west  of  Mar- 
quette,  was  probably  a  large  meteor,  and  a  similar 
suggestion  has  been  made  in  regard  to  the  iron  mount- 
ain in  the  State  of  Missouri. 

PAGE  77. 

6.  The  friction  of  the  water  produced  by  moving 
against  the  banks  and  bed   of  the  stream  produces 
heat,  which,  escaping  into  the  surrounding  air,  modi- 
fies its  temperature. 


TO  ELEMENTARY  MECHANICS.  79 

7.  First  by  the  heat  due  to  the  friction,  and  second, 
by  inducing  a  quicker  circulation  of  the  blood  more 
heat  is  supplied  to  the  parts. 

PAGE  77. 

8.  $2.50.     Tliis  exercise  is  intended  to  draw  atten- 
tion to  the  fact  that  the  value  of  wood  for  fuel  de- 
pends upon  its  capacity  for  producing  heat ;  or,  in 
other  words,  of  its  inherent  heat  energy.    If  the  heat- 
ing power  of  a  given  weight  of  hickory  be  100,  it  has 
been  found  that  the  heating  power  of  the  same  quan- 
tity of  oak  will  be  about  80,  and  of  maple  about  50. 
The  practical  value,  however,  of  fuel  may  be  governed 
largely  by  other  circumstances.     Thus,  when  a  fire  is 
wanted  for  only  a  short  time,  as  a  kitchen  fire  in  mid- 
summer, or  where  steam  must  be  raised  quickly,  etc., 
the  cheaper  fuel  may   be  quite  as  valuable   as  the 
more  costly. 

PAGES  78,  79. 

ARTS.  122-125. — Momentum,  according  to  New- 
ton's definition,  is  strictly  quantity  of  motion.  He 
says  (Principia  B.  1,  Def.  II.)  "  The  whole  motion 
is  the  sum  of  all  its  parts,  and  therefore  in  a  body, 
double  in  quantity,  with  equal  velocity,  the  motion  is 
double  ;  with  twice  the  velocity  it  is  quadruple."  It 
is  said  in  the  text  that  quantity  of  motion  does  not 
fully  express  the  desired  meaning,  but  this  is  due  sim- 
ply to  the  fact  that  quantity  had  not  been  defined. 
Including  Newton's  definition,  it  does  express  the 
meaning  correctly  and  fully. 

If  a  force  of  constant  intensity  acts  upon  a  free 
body  moving  from  rest,  the  product  of  the  force  and 


80  KEY  AND  SUPPLEMENT 

time  equals  the  momentum  produced.  Space  is  here 
entirely  abstracted  from  force  and  time.  Although 
the  body  cannot  move  without  involving  space,  yet 
all  considerations  of  space  must  be  discarded.  It  is 
immaterial  whether  the  space  over  which  the  body 
must  move  in  acquiring  the  velocity  v  be  great  or 
small ;  and  hence,  so  far  as  the  momentum  is  con- 
cerned, the  space  may  vanish. 

n\ 

From  the  equation  Ft  —  Mv,  we  have  F  —  M— , 

t 

v  f 

where  —  (or  for  a  variable  force,  we  have  in  the  nota- 
t  \ 

tion  of  the  calculus  — )  is  the  rate  of  change  of  the 
dtj 

1) 

velocity,  and  hence,  M  —  is  the  rate  of  change  of  mo- 

u 

mentum,  which  is,  according  to  the  second  law,  the 
measure  of  the  force  F.  We  thus  reproduce  the  ex- 
pression for  that  law. 

The  expression  Ft  is  not  the  momentum,  but  sim- 
ply its  equal  under  the  restrictions  given  above.  It 
has  been  proposed  to  give  a  special  name  to  this  prod- 
uct, just  as  7%  is  called  the  measure  of  work,  while 
its  equal  £jfy2,  in  case  of  a  free  body,  is  called  energy. 
Maxwell  called  Ft  an  impulse  (Matter  and  Motion, 
p.  44),  to  which  we  do  not  object,  since  the  effect  will 
be  the  same  whether  it  be  produced  in  an  impercep- 
tibly short  time,  or  in  a  longer  time.  But  the  prod- 
uct has  no  meaning  except  where  the  force  moves 
a  body.  If  F  be  a  mere  stress,  like  the  pressure  of  a 
stone  upon  the  earth  where  no  motion  is  involved, 
then  Ft  produces  nothing.  The  time  effect  of  a 


TO  ELEMENTARY  MECHANICS.  gl 

stress,   is   its  effect   in    moving  a    body  —  producing 
velocity. 

Much  has  been  written  in  regard  to  force,  vis  viva 
(now  called  energy)  and  momentum.  For  many  years, 
in  the  earlier  history  of  the  science  of  mechanics,  there 
were  long  and  sharp  discussions  as  to  whether  work 
or  momentum  was  the  proper  measure  of  force  ;  but, 
as  we  have  shown  in  what  precedes,  neither  is  the 
proper  measure,  and  hence  theirs  was  merely  a  war  of 
words.  One  factor  in  the  product  Fs  is  the  force  ; 
and  one  factor  in  the  product  Ft  is  also  force.  The 
former  placed  equal  to  j-3/v2,  and  solved  gives 


F  =  ?  -  :  hence  the  measure  of  force  is  the  rate  of 

s 

change  of  energy  per  unit  of  space;  and  we  have 
already  shown  that  it  is  also  the  rate  of  change  of  mo- 
mentum per  unit  of  time. 

Efforts  are  sometimes  made  to  determine  a  relation 
between  momentum  and  energy  ;  but  no  physical  re- 
lation exists,  and  hence  none  can  be  found.  In  order 
that  there  shall  be  a  ratio  between  them,  they  must 
have  a  common  unit.  Since  one  is  compounded  of 
force  and  time  in  which  space  is  excluded,  and  the 
other  of  force  and  space  in  which  time  is  excluded, 
they  have  not  a  common  unit. 

PAGE  79. 

We  here  note  the  three  following  statements  from 
different  authors. 

First,  In  Yan  Nostrand's  Engineering  Magazine  for 
1877  and  1878  is  a  series  of  articles  on  momentum, 
and  Vis  Viva  by  Prof.  J.  J.  Skinner.     On  pp.  129, 
4* 


82  KEY  AND  SUPPLEMENT 

130,  131,  it  is  stated  that  momentum  which  equals 
MV  represents  the  number  of  pounds  pressure 
which  the  mass  M  with  the  velocity  F  is  capable  of 
exerting  under  the  conditions  that  the  pressure  is  con- 
stant and  capable  of  bringing  the  body  to  rest  in  one 
second.  This  is  numerically  correct  as  a  deduction, 
but  in  the  articles  referred  to  there  is  apparently  a 
labored  effort  to  show  that  momentum  is  pressure 
only,  and  not  quantity  of  motion  (see  also  p.  137  of 
the  Eng.  Mag.}.  Also  on  page  132  it  is  stated  that, 
"  The  unit  of  momentum,  then,  is  a  force  of  pressure 
equal  to  one  pound  "  (see  also  p.  240).  In  this  defi- 
nition, the  element  of  time  does  not  appear,  but  it  is 
not  proper  to  drop  it  simply  because  it  is  one  second. 
The  above-named  writer  corrected  himself  in  a  later 
article,  page  501  of  the  same  Magazine. 

To  measure  anything  requires  a  unit  of  the  same 
kind  considered  as  a  standard.  Strictly  speaking,  the 
unit  of  momentum  is  the  momentum  of  a  unit  of  mass 
moving  with  a  unit  velocity.  Momentum  cannot  be 
measured  by  pounds  only.  (See  also  article  by  the 
author,  Eng.  Mag.,  vol.  xviii.,  1878,  p.  33.) 

Second,  It  is  stated  that  Beaufoy  determined  that 
a  body  of  one  pound  weight,  with  a  velocity  of  one 
foot  in  a  second,  strikes  with  a  pressure  equal  to 
0-5003  Ib. ;  and  hence  to  find  the  pressure  produced 
by  the  impact  of  any  projectile,  we  have  the  general 
iorm\\\&,  pressure  =  0-5003 Wv*  (Silliman's  Physics). 
Now  we  assert  that  the  formula  is  false.  Admitting, 
for  the  sake  of  the  argument,  that  he  did  find  such  a 
result  when  the  projectile  struck  a  hard  body,  like  a 
piece  of  iron,  it  would  have  been  very  much  less 


TO  ELEMENTARY  MECHANICS.  83 

had  the  body  struck  been  more  yielding,  like  a  gas- 
bag, or  a  sack  of  loose  feathers,  and  so  a  great  range  of 
values  might  be  found  depending  upon  the  character 
of  the  bodies. 

Third,  Professor  Tait,  iu  an  interesting  lecture  upon 
force,  delivered  before  the  British  Association,  1876, 
sajrs  (see  Nature,  1876,  p.  462),  "  With  a  moderate 
exertion  you  can  raise  a  hundred  weight  a  few  feet, 
and  in  its  descent  it  might  be  employed  to  drive  ma- 
chinery, or  to  do  some  other  species  of  work.  But  tug 
as  you  please  at  a  ton,  you  will  not  be  able  to  lift  it ; 
and,  therefore,  after  all  your  exertion,  it  will  not  be 
capable  of  doing  any  work  in  descending  again. 

"  Thus,  it  appears,  that  force  is  a  mere  name,  and 
that  the  product  of  the  force  into  the  displacement  of 
its  point  of  application  has  an  objective  existence." 

"  Force  is  the  rate  at  which  an  agent  does  work  per 
unit  of  length"  .  .  . 

These  definitions  have  already  been  referred  to  on 
p.  20  of  the  Key,  and  the  remarks  there  made  will  be 
more  readily  understood  in  this  place,  after  having 
passed  over  energy  and  momentum.  Force,  funda- 
mentally, is  a  quantity  instead  of  rate  ;  just  as  inter- 
est is  a  quantity — an  amount  of  money — and  not  rate. 
It  is  true  that  the  amount  of  money  paid  for  the  use 
of  a  hundred  dollars  is  identical  with  the  rate  per 
cent.,  but  every  one  readily  distinguishes  between 
rate  per  cent,  and  interest.  Professor  Tait  made  use 
of  interest  in  an  illustration  of  this  principle,  but,  we 
think,  used  it  improperly. 


Si  KEY  AND  SUPPLEMENT 

PAGE  83. 

ART.  131.—  If  A  =  Z,  and  k  =  1,  we  have  E  =  Fy 
hence  —  as  a  deduction  —  the  coefficient  of  elasticity 
may  be  defined  as  the  force  (stress)  necessary  to 
elongate  a  prismatic  bar  whose  section  is  unity  to 
double  its  length,  provided  the  original  conditions 
remain  constant  except  that  of  length  ;  that  is,  the 
elasticity  and  the  cross  section  must  both  remain  con- 
stant. But  these  conditions  are  never  realized,  hence 
this  definition  is  highly  ideal.  In  fact,  the  coefficient 
of  elasticity  is  constant  for  any  material  for  an  elonga- 
tion of  only  a  very  small  fraction  of  the  length  ;  but 
even  with  this  limitation  it  is  of  untold  importance 
in  certain  physical  sciences. 

PAGE  90. 

ART.  138.  —  Substituting  from  equations  (1)  and  (2), 
page  89,  we  have 

Mv?  +  M'  v\*  = 

~Mv  +  M'v  eM'      . 


,         ,N 

("- 


,,  r 

(  M  +  M  )2 

-  ZeM^vv   +  ZeMMvv'  + 


or 


(M+M1} 


TO  ELEMENTARY  MECHANICS.  85 

^vv'  + 


JTV2].  (2) 

Similarly, 


(3) 
Adding  equations  (2)  and  (3)  we  have 


+  + 


86  KEY  AND  SUPPLEMENT 


(M  +  M')<?MM'tf  +  (M  +  M')  ZMM'vo' 
(M  +  M')  2e*MM'vv'  +  (M  +  M')  M'*v* 

(M 


M+  M 


MM' 


Adding  and  subtracting  JO/"'(02+  ?)'2)  we  have 


M< 


T 

M  +  M 


'     [(1  - 

'  L 


«')2. 


TO  ELEMENTARY  MECHANICS.  87 

SOLUTIONS  OF  EXAMPLES. 

PAGE  91. 

1.  5  x  5  =  25  Ibs.  sec. 

0    ,  __  Fl    _  9000  x  10  x  12 
I  x  26000000 


„     ^       Fl       2500  x  2  x  12       a.    AAR    OAA  ,, 
3.  J^=  —-=  =  -  —  r—  =  -  -  —  =  24,  446,  200  Ibs. 


very  nearly. 


v_  Wo  -  W'v'_  120  -  120 
W+  W  ~~         18 


5.  Eqs.  (1)  and  (2),  p.  89  of  text,  give 
Wv+  W'v'  eW 

*=  -w^rw'~  '  w+  w(v' 

20  x  100  +  50  x  40        x  50 


feet; 

,  _  20  x  100  +  50  x  40      \  x  20 
~W~  ^70" 

65f  feet. 

PAGE  92. 

6.  Here  «'  =:  0,  and  we  have 

2000      25 
*  =  -       ~       x  100  =  - 


88  KEY  AND  SUPPLEMENT 

,      2000     10 
«i  =  --  +        x  100  =  42f  feet. 


7.  Here  «'  =  —  40  feet,  hence 


2000  -  2000   i  x  50  x  140 

- 


T 

feet  per  second. 


KA 


^  ,  _  2000  -  2000      j  x  20  x  40  _  2Q  f 
per  second. 

8.  By  Art.  132,  p.  84,  we  have 

h' 
e=-h> 

where  h  is  the  height  of  fall,  and  h'  the  height 
of  rebound.  The  sum  of  an  infinite  decreas- 
ing progression  is 

_  first  term, 
1  —  ratio 


TO  ELEMENTARY  MECHANICS.  89 

9.  If    <?  =  !,*  =    00. 

e  =  b,  s  =  \h 
e  =  i,  «  =  }f  A 

e  =  0,  s  =  h. 

10.  Let  M\)Q  the  mass  of  one  of  the  bodies,  and  v 

the  velocity  of  the  impinging  one  ;  its  kinetic 
energy  will  be 


The  kinetic  energy  of  the  two  bodies  after  im- 
pact will  be  least  when  both  are  non-elastic  ; 
in  which  case  the  common  velocity  after  im- 
pact will  be  (Eq.  (1)  p.  86), 


and  the  kinetic  energy  will  be 


and  the  kinetic  energy  before  impact  will  be 
jnst  twice  that  after,  and  cannot  exceed  that 
ratio. 

ANSWERS    TO    EXERCISES. 

1.  Yes. 

2.  No. 

3.  No. 

4.  By  equation  of  work  =  energy. 

5.  They  will,  at  the  instant  of  impact. 

6.  Yes. 

7.  No. 

8.  No. 


90  KEY  AND  SUPPLEMENT 

9.  By  preventing  so  great  a  loss  of  energy.  When 
the  wheels  of  a  car  strike  "  dead  "  against  a 
rail,  it  batters  the  rail,  thereby  doing  an 
amount  of  work  which  is  lost  to  the  train ;  but 
the  springs,  when  in  action,  prevent  such  a 
dead  blow  in  the  first  place,  and  then  by  their 
reaction  restore  a  portion  of  the  energy  to  the 
train.  In  short,  if  all  the  parts  were  perfectly 
elastic  there  would  be  no  permanent  battering 
of  the  parts,  and  no  energy  would  be  lost  by 
the  impact. 
PAGE  94. 

Statics  is  a  limiting  case  of  dynamics,  in  which  the 
applied  forces  mutually  destroy  each  other,  and  leave 
the  body,  so  far  as  its  condition  in  regard  to  rest  or 
motion  is  concerned,  the  same  as  if  no  forces  were 
acting.  As  such,  its  principles  may  be  established  in- 
dependently of  motion.  Many  writers  hold  that  the 
conditions  of  equilibrium  should  be  determined  inde- 
pendently of  all  considerations  of  motion,  and  we 
have  accordingly  given  the  usual  proofs,  although 
the  Newtonian  method,  given  in  Art.  52  of  the  text? 
is,  to  us,  quite  satisfactory. 

We  consider  that  the  confirmation  of  the  results, 
flowing  from  the  parallelogram  of  forces,  is  a  stronger 
confirmation  of  its  truth  than  any  formal  demon, 
stration  ever  made.  Nearly  all  the  formulas  of  me- 
chanics are  founded  upon  it.  The  principles  of  mech- 
anism, and  the  places  of  planets  and  comets  all 
involve  it.  Such  a  proposition  might  be  assumed 
without  proof,  and  its  truthfulness  be  confirmed  by 
its  leading  to  results  confirmed  by  daily  experience. 


TO  ELEMENTARY  MECHANICS.  .         91 

Formerly  a  proof  was  considered  so  important,  that 
many  different  methods  were  devised,  and  one  work 
gave  forty -five  different  proofs  of  the  parallelogram 
of  forces. 


SOLUTIONS    OF    EXAMPLES. 


PAGE  99. 

1.  Let  P  be  one  stress  and  F  the  other,  then  we 
have  (Eq.  on  p.  97  of  text), 


It  =  VP2  +  F*  +  ZPFcos  90°  =  VP2  +  F*. 
If  0  =  0,  then 


R  =  VP*  +  **  +  2PJP'  =  P  +  F. 
If  0  -  180°,  then 


R  =  VP2  +  ^'2  -  2PF=P  -  F. 


2.  Here  the  force  5  reversed  will  be  the  resultant 

of  the  other  two  ;  then 

52=32  4-  42+  2.3-4  cos  0; 
25-9-16      n 
~24"  ; 

.-.  0  =  90°. 

3.  We  have 


.'.  COS  0  —    —  \  • 

.-.  e  =  120°. 


92  KEY  AND  SUPPLEMENT 

4.  We  have 

It*  =  1002  +  1002  +  2  x  100  x  100  cos  60C 
But  cos  60°  =  |,  hence 

722  =  3  x  1002; 
.-.  R  -  100  V3. 

5    We  have 

(P  +  FY  =  P2  +  F*  +  2PPcos  B\ 
or, 
P2  +  2P^7  +  P2  =  P2  +  P2  +  2P^  cos  6; 

/.  cos  6>  =  1, 
and 


6.  We  have 

P2  =  (P  -  F)2  =  P2  +  F*  +2PF  cos  0, 

,     or, 

P2  -  2PF+F*  =  F*  +  F2  +  2PF  cos  ^; 

.-.  cos  0  =  -  1  ; 
.-.  0  =  1£0D. 

7.  From  the  proportion  on  p.  98  of  text  we  have 

50  :  F  :  :  sin  (F,R)  :  sin  115°, 
50  :  E  :  :  sin  (F,R)  :  sin  35°, 


TO  ELEMENTARY  MECHANICS.  93 

F:  E  :  :  sin  115°  :  sin  35°, 
P  :  7?  :  :  sin  30°  :  sin  35°. 

From  these  we  have 


sm  35 


R  =  !!  x  50  =  57.35  lbg> 

sin  30 


•    /  T-T  r>\      "'J  ^  sin  oo       ~  K~~,v 
sin  (JP,72)  =  —  =  0-5000  ; 


.-.  angle  (F,K)  =  30°. 

8.  The  string  will  make  a  right  angle  at  the  point 

where  the  weight  is  applied,  and  the  sides  of 
the  triangle  representing  the  forces  will  be  as 
3  to  4  to  5  ;  hence  we  have 

5  :  4  :  :  20  :x  =  16; 

5  :  3  : :  20  :  x  =  12. 
PAGE  100. 

9.  The  parallelogram   representing  the  forces  will 

be  a  rectangle,  of  which  the  diagonal  will  be 
a  diameter  of  the  circle. 

ANSWERS   TO    EXEECISE8. 

1.  It  will  be  the  resultant  of  two  forces  acting 
away  from  C,  one  of  which  will  equal  GA,  the 
other  CD  =  AS. 


94  KEY  AND  SUPPLEMENT 

2.  A  line  through  A   equal  and  parallel  to  a  line 

joining  J^and  B. 
3    They  would  not. 

4.  When  acting  upon  the  same  particle,  in  opposite 

directions,  and  equal  in  magnitude. 

5.  "With  4,  5,  and  9  they  can,  if  4  and  5  act  opposite 

to  9.  But  forces  3,  4,  and  8  cannot,  since  two 
of  them,  3  and  4  together,  do  not  equal  the 
third. 

6.  It  will  not.     The  resultant  takes  the  place  of 

the  other  two. 

PAGE  102. 

ART.  158.  It  will  be  observed  that,  in  Fig.  45,  ft 
is  the  complement  of  a  /  hence  cos  /?  =  cos 
(90  —  a)  —  sin  a.  hence  the  equations  for  ^Tand 
Y  become 

X.  =  FI  cos  <*!  +  Ft  cos  afz  +  etc.  =  0, 
Y  =  Fl  sin  az  +  FZ  sin  a2  +  etc.  =  0 ; 

from  which  we  see  that  the  equation  for  Y 
may  be  deduced  directly  from  that  of  X  by 
writing  sin  in  place  of  cos  in  the  first  equa- 
tion. 

SOLUTIONS  OF  EXAMPLES. 

PAGE  103. 

1.  We  have 

X  =  20  cos  30°  +  30  cos  90°  +  40  cos  150° 
50  cos  180°  =  R  cos  a, 


TO  ELEMENTARY  MECHANICS.  95 

Y=  20  sin  30°  +  30  sin  S0°+  40  sin  150°  +  50 
sin  180°  =  R  sin  a  • 

in  which  all  the  terms  are  written  as  positive, 
and  their  essential  signs  made  to  depend  upon 
the  trigonometrical  functions. 
Reducing  gives 

20  x  % V3  +  0  -  40  x  |  V3  -  50  =  R  cos  a, 

10  +  30  +  20  +  0  =  R  sin  a; 
or 

—  67-32   +  =  R  cos  or, 

60  =  R  sin  a. 
Dividing  gives 

Rco&a  _        67-32 

72  sin  a:  ~          60     ' 

or 

cota  =     -1-122; 
.-.  a  =  138°  17'. 

Squaring  and  adding  gives 

^(cos2«  +  sin2*)  =  (6T-32)2  -I-  (60)2. 
But  cos2  a  +  sin2  a  —  1 ; 
/.  7?  =  V8131-98 
=  90-18  Ibs. 


96  KEY  AND   SUPPLEMENT 

2.  7?  cos  a  =  20  cos  180°  +  10  cos  270°, 
R  sin  a  =  20  sin  180°  +  10  sin  270°  ; 

hence 

It,  cos  a  =  —  20, 
It  sin  or  =  —  10. 

Squaring  and  adding  gives 
R  =  V500  =  22-36. 

3.  We  have 

ltcosa  =  P  cos  0°  +  P  cos  90°  +  P  cos  225° 

+  P  cos  270°, 
E  sin  a  =  P  sin  0°  +  P  sin  90°  +  P  sin  225C 

+  P  sin  270°  ; 
hence 

R  cos  a  =  P  (1  -  £<v/2  ), 


=  -  1  V2P. 
Dividing  the  second  by  the  first  gives 


-24162  + 
.-.  a  =  292°  30'. 


TO   ELEMENTARY   MECHANICS.  97 

Squaring  and  adding  gives 

in2  a  +  cos2  a)  =  P2(l  -  J-\/2)2  +  P2(-  i  V2)2  J 


or 


R*  =  P2(2  -  V2) 
=  0-58579P2 ; 
.•.7£=0-765P. 


PAGE  104,  ART.  162. — A  single  force  whose  line  of 
action  does  not  pass  through  the  centre  of  a  free 
body,  produces  rotation  as  well  as  translation.  The 
measure  of  the  effect  of  a  force  in  producing  rotation 
is  proportional  to  the  moment  of  the  force;  as  is 
shown  from  the-  fact  that  the  moment  is  proportional 
to  the  work  done  by  the  force. 

The  theory  of  moments  is  here  discussed  without 
reference  to  the  bodies  upon  which  the  forces  act. 

PAGE  110,  ART.  176.— The  cut,  Fig.  55,  should  be  as 
here  given ;  that  is,  in  the 
typical  figure  the  force  should 
be  positive  away  from  the 
origin,  and  so  placed  that  it 
would  produce  positive  rota- 
tion about  the  origin,  O,  and 
the  angles  a  and  ft  be  acute, 
as  shown  in  Fig.  42  of  the 
text ;  so  that  the  signs  of  the  terms  in  the  analytical 
expression  for  the  moment  will  flow  directly  from 
the  trigonometrical  functions.  This  being  done  the 
5 


98  KEY  AND  SUPPLEMENT 

typical  form  of  the  expression  for  the  arm  of  the 
force  will  be 

Oa  =  x  cos  ft  —  y  cos  a. 

In  the  old  Fig.  55,  the  angle  between  the  axis  of  x 
and  the  direction  line  of  the  force  was,  as  shown 
in  Art.  157, 

a  =  180°  +  dOb ; 
and  similarly, 

/S  =  180°  +  Y0d\ 
.•.  cos  a  =  —  cos  dObj 

cos  ft  =  —  cos  YOd, 
which  values  give  for  the  arm 

Oa  =  y  cos  a  —  x  cos  /?, 

as  given  in  the  text.  We  call  the  former  value  the 
typical  one,  and  the  latter  a  deduced  one.  The  for- 
mer should  always  be  used  in  connection  with  the  true 
value  of  the  angles  a  and  /3. 


SOLUTIONS  OK  EXAMPLES. 

PAGE  115. 

L  We  have 


•' 

in  which 

W=  20  Ibs., 

AC  =  24  inches,  AS  =  6  inches,  AD  —  4 
inches. 


TO  ELEMENTARY  MECHANICS.  99 

To  find  AE  ';  in  the  right-angled  triangle  DAB 
we  have 

A  D 
tang  B  =  ±       =  i  =  0-666  +  ; 


.-.  B  =  33°  41'. 
Then 

AE  =  AB  sin  B 

=  3-327  +  inches. 

Substituting  above  gives 
24  x  20 


2.  Taking  the  origin  of  moments  at  D  we  have 
P  —  ^ 

~ 


20  x  24      10A  ., 
—  :  -  =  120  Ibs. 


3.  Taking  the  origin  of  moments  at  B  we  have 

W.BC^F.AB, 
or 


PAGE  116. 

4.  If  t  =  TF,  we  have 


100  KEY  AND  SUPPLEMENT 


or 

=  AC. 


From  the  right-angled  triangle  AEB  we  have 


sin  45C 
substituting, 

AC 


5.  Taking  the  origin  of  moments  at  A  we  have 

t  x  AC=  BE  x   W. 
But  from  the  example 
DB=2AB,6  =  45°,  TF=  £0  Ibs. 
From  the  figure 

BE-  AB  sin  45° 

=  $V2  AB  =  0-7071AB. 

BE     AB  sin  45° 


.-.  9>  •  =  20°  42'  17". 
To  find  A  C  we  have  from  the  figure 

AC=AB&\nABC 

=  J#  sin  (0  -  cp) 


TO  ELEMENTARY  MECHANICS.  101 

=  AB  sin  24°  17'  43" 
=  04115  AB. 

Substituting  in  the  first  equation  above  gives 

Q.7071  x  50 
04115 

=  85-97  +  Ibs. 

To  get  the  compression  on  the  bar,  take  the  origin 
of  moments  at  D,  in  which  case  the  moment  of 
the  tension  will  be  zero.  The  perpendicular 
from  D  upon  AB  prolonged  will  be  AD  sin  #, 
and  from  D  perpendicular  upon  the  vertical 
through  B  will  equal  BE  =  AB  sin  6  ;  hence 
we  have  —  calling  c  the  compression  — 

c.AD&hie  =  W.AB&inff; 

AS  ^ 

~  AD 

To  find  AD  we  have 

sin  ABD  :  sin  cp  :  :  AD  :  AB  ; 


/in   _  04115 

.*.  A.JJ  —  — 


AB 


0-8592  +  ' 
which  substituted  above  gives 

c-  0-8592  W 
=  42-96  Ibs. 


102  KEY  AND  SUPPLEMENT 

PAGE  116. 

6.  Let  fall  a  perpendicular  p  from  B  upon  AC,  then 

p  —  AB  sin  A. 

Let  D  be  directly  under  W,  then  taking  the 
origin  of  moments  at  13  we  have 
t.p  =  W.BD; 

BCcos  CBD 


AB  sin  A 


W. 


l.lit  =  W,  then  AB  sin  BAG  =  BD,  or  BC will 
bisect  the  angle  ACD. 

8.  Take  the  origin  of  moments  at  ^4.  Let  fall  a 
perpendicular  from  ./4  upon  CB  produced;  its 
length  will  be 

p  =  AB  sin  CBD 

Q 

=  6x 
=  6  x 


BG 

8 


48 

feet. 


~  8-9442 

Let  c  be  the  compression  on   BC,  then   the 
equation  of  moments  becomes, 

c.p=   W.  AD; 
AD 


P 
_  10  x  8-9442 

48 
=  931 -Tibs. 


500 


TO  ELEMENTARY  MECHANICS.  103 

9.  For  equilibrium  we  have 

Ol  =  -^ — ^-  be 


0 

=  co; 
or  there  can  be  no  equilibrium. 

10.  We  will  have 

be  =  P*~Pl  Ob 

_2  x  0 

Pi 
=  0; 

or  the  forces  must  act  at  the  same  point. 
PAGE  118,  ART.  187. — It  is  well  to  illustrate  this  article 
still  further.  If  the  forces 
constituting  the  couple  be  & 
equidistant  from  the  cen- 
tre c  of  the  body,  it  is  suffi- 
ciently evident,  without  a  thorough  demonstration, 
that  it  will  produce  rotation  only.  But  is  it  equally 
evident  that,  if  the  same  couple  act  upon  the  same 
body  in  such  a  way  that  the  <p 

points   of    application    are 

both  on  one  side  of  the  cen-  ~r~ — 

tre,  it  will  produce  rotation 
only,  and  the  same  amount 
of  rotation  as  in  the  preceding  case  ?  According  to 


104  KEY  AND  SUPPLEMENT 

the  proposition  it  will,  and  we  will  prove  it  by  a  special 
solution. 

At  a  point  d,  such  that  do  =  cb,  introduce  two 


-9 

P 

P 

equal  and 

HP            HP 

T. 

o  p  p  o  s  i  t  e 

A         \ 

,        N 

/            \ 

f                   0 

B    .,                    , 

7 

6 

v     C 

a 

loi  ces,  eacn 

4 

HP           HP 

4 

equal  to|P, 

9 

P                       5IP 

and  call  the 

upper  |P,  1,  a*nd  the  lower  half  4.  Since  these  are 
in  equilibrium  the  problem  will  be  the  same  as  before. 
Separate  the  force  at  b  into  two  equal  parts,  and  for 
the  sake  of  convenience  call  one  part  2  and  the 
other  5.  Now  the  resultant  of  4  and  5  will  be  a 
force  equal  to  their  sum,  or  P,  applied  at  the  centre 
c  —  which  force  call  9.  Combining  1  with  2  we  have 
a  couple  whose  arm  is  db,  and  the  moment  will  be 
%P.db  =  P.cb.  Similarly  at  e,  at  a  distance  ec  = 
ca,  introduce  two  equal  and  opposite  forces,  each 
equal  to  |P  ;  and  separate  the  force  at  a  into  two 
equal  parts.  Combining  %P  at  a  with  the^P  above 
e  gives  a  resultant  equal  to  P  applied  at  c  acting 
down,  and  marked  —  9,  which  will  equilibrate  +  9, 
and  there  will  be  ixo  motion  of  translation.  There 
will  remain  the  couple  ^P.ea,  which  will  produce 
rotation  only  about  the  centre  c.  Finally,  the  two 
couples  %P.db,  and  iP.ea,  each  producing  rotation 
about  the  centre  c,  but  in  opposite  senses,  arc  equiva- 
lent to  the  single  couple. 


iea)  =  P(cb  -  ca)  =  P.ab  ; 
henoe,  the  body  will  rotate  about  its  centre  of  gravity, 


TO  ELEMENTARY  MECHANICS.  1Q5 

and  the  rate  of  rotation  will  be  independent  of  the 
points  of  application  of  the  forces,  and  dependent  only 
upon  the  moment  of  the  couple. 

PAGE  120,  ART.  190. — This  article  contains  all  the  prin- 
ciples of  the  simple  lever.  In  some  works  levers  are 
divided  into  three  classes,  but,  mechanically,  there  is 
no  distinction  between  them.  It  is  only  necessary 
for  equilibrium  that  the  sum  of  the  moments  be  zero 
— or  that  the  moments  of  the  forces  which  turn  the 
lever  one  way  equals  the  moments  of  those  which 
tend  to  turn  it  the  opposite  way. 

PAGE  121. 

ART.  192. — The  force  F'  and  D  produces  equilib- 
rium in  the  system  ;  hence  a  single  force  equal  and 
opposite  to  F  at  D  will  produce  the  same  effect  as 
the  three  forces  F,  P,  and  P. 

ART.  197. — Assume  that  any  number  of  forces  act 
upon  a  body  in  any  manner;  they  may  produce  both 
translation  and  rotation.  The  measure  of  their  effort 
to  produce  rotation  will  be  the  sum  of  their  moments, 
wherever  be  the  origin,  and  the  sum  of  these  mo- 
ments will  be  equivalent  to  a  single  couple,  Arts.  184 
and  18G.  Hence,  if  there  be  no  effort  at  rotation,  the 
sum  of  the  moments  will  vanish  for  any  and  every 
point  assumed  for  the  origin  of  moments.  The  only 
other  tendency  to  motion  is  that  of  translation  ;  in 
which  case  there  will  be  a  single  resultant  passing 
through  the  centre  of  the  body.  If  there  be  a  result- 
ant, the  moment  will  be  zero  when  the  origin  of  mo- 
ments is  on  the  line  of  the  resultant,  Art.  189  ;  and 
will  have  a  finite  value  when  the  origin  is  not  on  the 


106  KEY  AND  SUPPLEMENT 

line  of  the  resultant ;  and  if  there  be  no  resultant  the 
latter  moment  will  also  vanish. 

SOLUTIONS  OF  EXAMPLES. 
PAGE  125. 

1.  For,  according  to  the  triangle  of  forces,  the  re- 
sultant of  two  of  them  will  equal  in. magnitude  that 
represented  by  the  third  side,  but  its  direction  of  ac- 
tion will  be  opposite  to  that  represented   by  the  third 
side,  and  at  a  distance  from  it  equal  to  the  altitude  of 
the  triangle.     See  Fig.  71  of  the  text,  only  in  this 
case  the  third  force  will  wAfrofan,  A  towards  B. 

2.  Inscribe  a  circle  in  the  triangle ;  then  will  the 
radius  r  be  the  common  arm  of  the  three  forces  in 
reference  to  the  centre  cf  the  circle,  and  we  will  have 
the  equation  of  moments 

It.r  =  P.r  +  F.r; 
.-.  R  =  P  +  P. 

3.  The  point  of  application  of  the  resultant  of  two 
of  them  will  be  at  the  middle  point  of  the  side  of  the 
triangle  between  them ;  and  the  point  of  application 
of  this  resultant  and  the  third  weight  will  be  at  two- 
thirds  the  distance  from  the  third  weight  on  the  line 
joining  them  ;  and  this  will  be  the  required  point.   It 
is  at  the  intersection  of  the  medians  of  the  triangle 
(Art.  222). 

4.  Let  P  and  Fbe  the  respective  amounts;  then 
taking  the  origin  of  moments  at  the  weight  we  have 
the  relative  moments  2.P  and  U^;  hence 

2P  =  F. 

Also,  since  the  sum  of  P  and  F  equals  the  entire 
weight, 


TO  ELEMENTARY  MECHANICS.  1Q7 

P  +  ^= 


Substituting, 

3P  =  175  pounds  ; 
.-.  P  =.  58|  pounds, 
and 

jF  =  116f  pounds. 

5.  Let  W  =  500,  the  arm  of  which  in  reference 
to  the  point  B  will  be  one-half  of  AS,  or  1  foot  ; 
and  the  arm  of  F  will  be  DB  =  3  feet  ;  hence  the 
equation  of  moments  will  be 


=  500  Ibs. ; 
.-.  jF=166flbs. 

6.  Let  x  =  the  required  distance,  then  will  the  lever 
arm  of  the  weight  be  x,  and  of  the  man  8  —  a? ;  hence 
we  have,  taking  the  origin  of  moments  at  the  fulcrum, 

175  (8  -  x)  =  4000ss, 
or 

(4000  +  175)*  =  8  x  175; 
/.  x  =  H$£  feet 
=  4ffa  inches. 

ANSWERS  TO  EXERCISES. 
PAGE  126. 

1.  It  is.     Foot-pounds  of  rotary  effort. 

2.  The  resistance    in   pounds   which   is  overcome 

through  a  certain  number  of  linear  feet. 

3.  It   is  the  momentum   of    a  given    number  of 


108  KEY  AND  SUPPLEMENT 

pounds  of  mass  moving  at  a  given  rate  in  feet 
per  second. 

4.  This  question   is  defective,  because  velocity  in- 

volves a  unit  of  time,  which  may  be  one  sec- 
ond, one  minute,  or  any  other  unit.  Assum- 
ing that  the  velocity  is  feet  per  second,  the 
unit  will  be  1  pound  of  mass  x  1  foot  per 
second  x  1  foot  for  the  arm. 

5.  It  can,  and  will  always  do  so  in  a  free  body  if 

the  line  of  action  of  the  force  does  not  pass 
through  the  centre  of  the  mass. 

0.  They  cannot.  Since  neither  couple  acting  sepa- 
rately can  produce  translation,  they  cannot 
produce  it  when  acting  together.  The  resultant 
of  two  couples  is  a  single  couple,  Arts.  185  and 
186,  and  for  this  reason  can  produce  rotation 
only. 

7.  It  will  be  100  Ibs.  more.    Pulling  down  with  his 

hands  will  add  nothing  to  the  pressure  of  his 
feet,  for  that  effort  is  resisted  by  an  equal  up- 
ward push  of  his  shoulder.  A  man  by  pulling 
upward  on  the  straps  of  his  boots,  does  not, 
thereby,  diminish  the  pressure  of  the  boot 
upon  the  floor,  although  it  increases  the  pressure 
between  his  foot  and  the  boot.  No  "perpet- 
ual motion  man  "  has  yet  thrown  himself  over 
a  fence  by  pulling  on  the  straps  of  his  boots. 

8.  If  the  cutting  is  uniform  it  is  ;  but  if  the  timber 

at  one  side  of  the  hole  is  harder  than  at  the 


TO  ELEMENTARY  MECHANICS.  1Q9 

other,  there  will  be  a  side  push,  tending  to 
force  the  auger  out  of  line. 

9.  It  will. 

SOLUTIONS  OF  EXAMPLES. 
PAGE  133. 

1.  Let  R  be  the  resultant,  and  x  the  distance  of  its 

point  of  application  from  the  force  6.  Take 
the  origin  moments  at  6 ;  then  we  have  the 
equation  of  moments 

Ex  =  11  x  5, 

and  of  forces 

R  =  6  +  11  =  17; 

.-.  x  =  -f 4  =  3,4T  feet. 

2.  We  have,  using  the  same  notation  as  before, 

fix  =  11  x  5  =  55, 

R  =  11  -  6  =  5  ; 

...  x  =  jyi  =  11  feet. 

3.  Take  the  origin  of  moments  at  the  extremity  of 

the  line  near  the  weight  2,  and  retaining  the 
same  notation  as  above,  we  have 

Rx  =  2-2  +  3-3  +  4-4  +  5-5  =  54 

^^2  +  3+4+5=14  Ibs. ; 
.-.  x  =  54  -r- 14  =  3f  feet. 

4.  Take  the  origin  of  moments  at  A,  then  we  have 


HO  KEY  AND  SUPPLEMENT 

Ex  =3x0  +  4x3  +  5x7  +  6x5  =  77; 

J?=3  +  4  +  5  +  6  =  18  Ibs.; 
/.  x  —  77  -f-  18  =  4  feet  3£  inches. 
PAGE  134. 

5.  Let  x  be  distance  from  A  where  P  is  applied, 
to  the  point  of  application  of  the  resultant  ; 
take  the  origin  of  moments  at  the  point  of  ap- 
plication of  the  resultant,  then  will  the  arm 
of  P  be  *  sin  cp,  and  of  F,(x  +  AB)  sin  cp,  as- 
suming P  >  F\  hence 
Px  sin  <p  =  F(x  +  AB}  sin  cp  ; 


which  substituted  in  the  preceding  equation 
gives 

PF  PF     ,n  . 

p  _  p  AB  sin  <p  =  p  _  p  AB  sm  cp. 

ANSWERS  TO  EXERCISES. 
PAGE  134. 

1.  It  has  not.     We  may  say  that  it  has  one  at  in- 

finity, which  is  equivalent  to  saying  that  it  has 
none. 

2.  No.     The  sum  of  the  forces  must  also  be  zero. 

If  the  sum  of  the  moments  in  reference  to 
three  arbitrary  points  is  zero,  they  will  be  in 
equilibrium  (see  Art.  195  of  the  text). 

3.  When  they  form  a  couple. 

4.  Because  the  resultant  is  zero. 

5.  It  will. 

6.  It  will. 


TO  ELEMENTARY  MECHANICS. 

7.  It  may  vary  directly  as  the  distance  from  the 
centre ;  or  inversely  as  the  distance  ;  or  as  any 
power  of  the  distance  ;  or  any  root  of  the  dis- 
tance ;  or  as  any  power  for  a  part  of  the  dis- 
tance, and  any  root  for  the  remaining  distance  ; 
or  in  any  other  way,  provided  the  concentric 
shells  comprising  the  sphere  shall  be  of  uni- 
form density. 

PAGE  135. — Many  of  the  properties  of  the  centre  of 
gravity  were  developed  as  long  ago  as  in  the  days  of 
Archimedes.  The  properties  are  not  only  of  great 
importance  in  statics;  but  when  the  principles  of 
dynamics  were  developed — after  Galileo's  time — they 
were  found  to  be  no  less  important  in  that  science. 
We  mention  only  one — A  free  rotating  body  rotates 
about  an  axis  through  its  centre  of  gravity  ;  or,  more 
strictly,  through  the  centre  of  the  mass. 

ANSWERS  TO  EXERCISES. 
PAGE  138. 

1.  The  vertical  through  the  centre  of  gravity  of 

the  carriage  and  load  must  intersect  the  ground 
between  the  wheels.  In  order  that  it  shall  over- 
turn, the  vertical  must  fall  outside  the  base  of 
the  wheels.  A  carriage  in  motion  may  over- 
turn when  it  would  not  if  standing,  on  account 
of  the  inertia  of  the  mass.  A  sudden  side 
"lurch"  may  induce  a  rotary  movement  suffi- 
cient to  overturn  it. 

2.  He  can  stand  so  long  as  the  vertical  through  the 

centre  of  gravity  of  his  body  falls  within  the 
base  occupied  by  his  feet. 


112  KEY  AND  SUPPLEMENT 

3.  Some  parts  of  his  body  must  move  backward. 

The  space  occupied  by  his  feet  being  of  finite 
size,  he  may  move  his  head,  or  other  parts  of 
his  body,  to  some  extent  without  endangering 
his  stability. 

4.  Because  the  base  being  so  very  narrow,  with  only 

two  legs,  a  small  displacement  will  cause  the 
vertical  through  the  centre  of  gravity  to  fall 
without  the  support. 

5.  Because  in  Fig.  77  the  centre  of  gravity  must 

be  moved  further  than  in  Fig.  78,  and  also 
must  be  raised  through  a  greater  height.  Some 
writers  consider  the  height  through  which  the 

d*  o 

centre  of  gravity  must  be  raised  in  order  to 
overturn  a  body,  a  measure  of  its  stability. 

6.  Because  the  line  through  the  point  of  support 

and  the  centre  of  gravity  of  the  book  will  be 
inclined  to  the  edges,  and  as  the  former  will 
be  vertical,  the  latter  must  be  inclined. 

7.  It  may.     Such  will  be  the  case  with  a  cylinder 

resting  on  its  convex  surface.     It  will  be  in 

indifferent  equilibrium  in  reference  to  rolling, 

but  stable  in  reference  to  a  longitudinal  mo- 
tion. 

8.  The  centre  of  gravity  is  below  the  top  of  the 

post. 

9.  In  both  these  exercises  the  balls  are  so  much 

heavier  than  the  bodies  to  which  they  are  at- 
tached, that  the  centre  of  gravity  of  the  whole 
device  is  below  the  support. 


TO  ELEMENTARY  MECHANICS.  H3 


PAGE  140. 

x  is  read  "  x  das/i" 


SOLUTIONS  OF  EXAMPLES. 

PAGE  140. 

1.  Let  I  be  the  length  of  the  line,  and  x  the  dis- 

tance of  the  centre  of  gravity  to  the  smaller 
weight ;  then  will  I  —  x  be  the  distance  to  the 
other,  and  the  equation  of  moments  gives 

1..X  =  n(l  —  x) ; 

n    j 

•  •  00  —  ^  t. 

1  +  n 

If  n  =  2,  then  x  =  \l. 
If  n  =  3,  then  x  =  %l. 

2.  The  middle  of  one  side  will  be  the  centre  of 

gravity  of  two  of  them,  and  the  centre  of 
gravity  of  the  three  will  be  in  the  line  joining 
this  point  with  the  vertex,  and,  according  to 
the  preceding  example,  it  will  be  at  two-thirds 
the  distance  from  the  apex.  Hence  the  centre 
of  gravity  will  be  at  two-thirds  the  distance 
from  any  apex,  on  a  line  drawn  to  the  middle 
of  the  opposite  side. 

3.  The   centre   of  gravity  of   the    weights    1  and 

2  will  be  in  the  line  joining  them,  and  at 
two-thirds  the  distance  from  1 ;  and  of  the 
three  weights  it  will  be  in  the  line  joining  the 
former  point,  and  the  weight  3,  and  at  its 


114  KEY  AND  SUPPLEMENT 

middle  point.  The  solution  is  the  same  whether 
the  triangle  be  equilateral  or  scalene. 
4.  The  centre  of  gravity  of  the  three  weights  at 
the  base  will  be  at  the  centre  of  the  base,  and 
of  the  four  weights  in  a  line  joining  the  apex 
with  the  centre  of  the  base,  and,  according  to 
the  first  example,  at  three-fourths  the  distance 
from  the  apex. 

ART.  217.  —  A  line,  in  mechanics,  is  a  body  from 
which  all  dimensions  are  abstracted  except 
that  of  length. 

PAGE  142,  ART.  220.  —  This  article  is  introduced  here  — 
in  advance  of  its  proof  —  partly  to  classify  it  with 
lines,  and  partly  to  furnish  exercises. 

SOLUTIONS   OF   EXAMPLES. 

1.  Join  the  centre  of  gravity  of  one  edge  with  that 

of  another,  and  the  centre  of  this  line   with 
the  centre  of  another  edge,  and  so  on. 

2.  Half  the  diagonal  of  the  base  (a  side  being  1) 

will  be  ^A/2,  and  the  length  of  one  of  the  lat- 


eral edges  will  be  Vl  +  (f^f  =  Vf  - 
There  are  four  edges,  and  hence  the  entire 
lengths  will  be  2V&  The  lengths  of  the 
sides  of  the  base  will  be  4.  The  centre  of 
gravity  of  the  four  lateral  edges  will  be  at  one- 
half  the  altitude,  and  of  the  edges  of  the  base 
it  will  be  at  the  centre  of  the  base.  Taking 

o 

the  origin  of  moments  at  the  apex,  and  x  the 
distance  of  the  centre  of  gravity  from  the  apex, 
we  have 


TO  ELEMENTARY  MECHANICS.  115 

(2\/6  +  4)  x  =  2V6  x  £  +  4  x  1  , 


3.  In  the  equation   in  Article  220,  make  BO  =  r, 
AC  =  2r  j  then  arc  ABC  =  nr,  and  we  have 


TTT         TT 
PAGE  143. 

4.  We  will  have  5(9  =  r,  ^£7  =  the  chord  of 
60°  =  r,  the  radius,  and  AB  =  2?rr  jfo  =  \irr  ; 
hence 

* 

Oc  = 


5.  We  will  have  BO  =  r,  AC  =  the  side  of  an  in- 
scribed square  =  r^/2,  and  ABO  =  the  arc  of 
a  quadrant  =  \nr  ;  hence 


ART.  221.  —  A  surface  is  a  body  from  which  all  di- 
mensions are  abstracted  except  length  and 
breadth. 

SOLUTIONS  OF  EXAMPLES. 
PAGE  145. 

1.  It  will  not,  for  the  moment  of  the  part  next  to 

the  apex  will  be  less  than  the  part  next  to  the 
base. 

2.  When  the  sides  adjacent  to  the  vertical  angle 

are  equal,  the  bisecting  line  will  pass  through 


116  KEY  AND  SUPPLEMENT 

the  centre  of  gravity  of  the  triangle.  In  other 
cases  it  will  not. 

3.  The  altitudes  will  be  the  same,  and  the  centre 

of  gravity  will  be  at  one-third  the  altitude  from 
the  base. 

4.  The  area  of  the  larger  circle  will  be  7tl&\  of 

the  smaller  nip;  and  of  the  remaining  part 
n^W1  —  i*).  Taking  the  point  A  for  the  origin 
of  moments,  we  have 


Rs  -  i* 

•         >3/»    -    _ 

~ 


-  ', 
+  Er  + 


T+r 

Iir  =  i  5,  then 

Ac  =  %R. 
Ifr=  R,  Ac  =$R. 

5.  Let  AF  =  AD  —  a  ;  then  the  diagonal  AE  = 
a  A/2,  CE  =  i#A/2,  and  the  equation  of  mo- 
ments will  be 


.-.  AS  =  &aV%  =f  i«A/2  =  $AC. 
PAGE  146! 

6.  The  centre  of  gravity  of  the  triangle  ABC  w\\\ 
be  at  \  of  CFirom  F\  of  DCE,  \  of  CG  from 
G;  or  from  ^it  will  be  FG  +  £  ^^  The  tri- 
angle ^£(7  will  be  to  triangle  DCE&s  (FCf 


TO  ELEMENTARY  MECHANICS. 

is   to   (£<7)2;   and   ACS:  ADEB  =  (FC)*: 


Taking  the  origin  of  moments  at  F  we  have, 
ADEB   x   Fg  =  ACS  x  $FC  -  DGE  x 

(FG  +  ICG}. 
From  the  proportions,  we  have, 

(FC?-(QC? 


Substituting, 

((FC'f  -  (GC}*)Fg  =  l(FCf  -  (CG)\FG 


From  the  figure  we  have 
CG  _DE 
FG  ~  AB  ' 
and 

FO=CG  + 
CG         _DE  . 
"CG  +  FG  ~  AB' 


hence  we  find 

AT) 


_ 
AB-DE 

Substituting  above  gives 

AB         \2        (         DE 

AB  -  BE)  '  (AS  -  BE 


FG. 

* 


17  v        n- 

K  '  x      ~ 


118  KEY  AND  SUPPLEMENT 


DE 


f     AB     V;  wr* 

(AB-DE)  (FG}  "  \AB-DE) 


Reducing  gives 

-(DEY  T      (AB)* 

(AB  -  DE)*     •?  '         T  i(AB  -  DE? 

(DE)\ZAB-WE)-] 
(AB  -  DE)*        J 

,        r(^g)8  -  3AB.(DE)*  +  2(Zl57n  . 
^L  (AB-DE)*  J' 

/.  Fg  = 


(  A  B  )3  - 


-  DE)(AB  -  DE)(AB  +  DE) 


-  ZAB.DE+  DE*)(AB  +DE)_\ 

AB  +  WE 
AB+  DE' 


7.  The  slant  height  will  be  A/V2  +  A2?  and  the  lat- 
eral area,  2?rr4vV2  +  A2;  and  the  centre  of 
gravity  of  the  lateral  area  will  be  in  the  axis 
at  \li  from  the  apex.  Let  x  be  the  required 
distance,  then 


+  TrrVr*  4-  A2)  #  = 


o  v  '    +  A2  4-  r  , 
/.  x  —      ,..,         A. 


TO  ELEMENTARY  MECHANICS. 


119 


PAGE  149,  ART.  230.  —  PROBLEM.     To  find  the 
of  gravity  of  a  segment  of  a 
sphere. 

A  segment  of  one  base,  GAH, 
is  considered  in  the  text. 

To  find  the  volume  of  the 
segment  AGH,  we  have  from 
geometry 


centre 


"We  put  this  under  another  form  —  thus,  in  the  right- 
angled  triangle  AlfC,  we  have 

(AH)2  =  (AC)2  -  (CH)Z 

=  (CG)Z  -  (CG  -  GH? 
=  2CG.GIf-  (Gil)* 
=  GH(2CG  -  GH), 

which,  substituted  in  the  preceding  expression,  gives 
for  the  volume  of  the  segment, 


which  is  the  value  used  in  the  text.  If  r  be  the  radius 
of  the  sphere,  and  h  the  altitude  of  the  segment,  the 
expression  becomes 

rch\r  -  lh). 

For  the  volume  of  the  spherical  sector  ACG,  we  have, 
from  geometry, 


120  KEY  AND  SUPPLEMENT 


x  GH 

=  f  7ZV'2A, 

which  is  also  the  value  used  in  the  text. 

The  volume  of  the  cone  the  radius  of  whose  base  is 
AH,  and  altitude  UC,  is 


n(AHJ-  x 
==*(/*-  (r  -  A)*H(>-  -  A) 


-  h)(r  -  li). 

These  values  in  the  first  equation  of  the  article 
gives 

r  , 


-  A)  -  3(r*  -  (r  -  Jif)(r  - 
" 


To  put  this  under  another  form,  let  the  angle 
ACG  =  6,  then 

h  =  r  —  CH  —r  —  r  cos  9  —  /•(!  —  cos  ff)  ; 
hence 

2r  —  h  =  2r  —(r  —  r  cos  0) 
=  2r  —  r  +  r  cos  0 
=  r  (1  +  cos  6). 


TO  ELEMENTARY  MECHANICS. 

From  trigonometry 

2  cos*J0  =  1  +  cos  0  ; 
.-.  (2r  -  /O2  =  ^  cos4  \B. 

Also  we  find 


=  r  —  l(r  —  rcosd) 
=  ^r(2  +  cos  6}  ; 


and  these  substituted  above,  give 
3  cos4  iff 


If  A  =  0,  we  have  0=0,  and  the  segment  will  vanish, 
and  we  have 

<V.=rt 
as  it  should. 
If  h  —  2r,  6  =  180°,  and  we  have 


hence  the  centre  of  gravity  of  the  sphere  will  be  at 
its  geometrical  centre,  as  it  should. 

For  the  hemisphere,  6  =  90°,  and  we  have 

<%'  =  f  UV2)4/1 


If  the  segment  has  two  bases,  let  B  be  the  angle 
subtended  by  the  radius  of  the  upper  base,  and  cp  the 
angle  subtended  by  the  radius  of  the  lower  base,  then 
by  taking  the  difference  of  the  moments  of  both  seg- 
ments, and  the  segment  on  the  upper  base,  we  find 
6 


122  KEY  AND  SUPPLEMENT 

Cg'  = 

(1  —  cos  <p)2.  cos4£<p  —  (L  —  cos  ff)2. 
(1  -  cos  (p?(*  +  cos  y)  -~(l  -  cos  0)2(2  +  cos  6}  r' 

SOLUTIONS  OF  EXAMPLES. 
PAGE  150. 

1.  In  Article  229,  CG  is  the  radius  of  a  circle,  and 
by  the  conditions  of  the  problem  ^ZTwill  also 
be  a  radius;  hence  we  have 

1(2<7£  -  Gil) 
=  |(2r-r) 


2.  Let  x  be  the  required  distance  to  the  centre  from 
the  common  tangent  point  ;  then,  according  to 
Article  202,  the  moment  of  the  difference  of 
the  spheres  will  equal  the  difference  of  their 
moments,  and  we  have 


JPr  +  Rr1  + 


IP  4-  Itr  +  i* 
If  r  =  0,  we  have 

x  =  R, 

and  the  centre  of  gravity  will  be  at  the  centre 
of  the  sphere. 
If  r  =  R,  we  have 


which   gives  the  centre  of  gravity  when  the 


TO  ELEMENTARY  MECHANICS. 


123 


thickness   of  the    solid,  opposite   the  tangent 
point,  is  infinitesimal. 

3.  To  find  the  versed-sine  Gil,  Fig.  94,  we  have 
in  this  case  CB  =  ?',  HE  z=  ±r ; 


.-.  cn=  v*1*  -Tv/<2= 

.-.  Gil  -(1  -f\/l5)r. 
In  the  equation  of  Article  230, 
Cg  =  |[2r  -  (1  - 
and  we  have 


12(1  -  ±Vl5)    x  y-2  x  [(r- 
t  V  ~ 


=  0-977  7-  nearly. 


4.  The  centre  of  gravity  will  be  £  of  8  inches,  or 
2  inches  from  the  base,  hence  the  line  joining 


124  KEY  AND  SUPPLEMENT 

the  point  of  suspension  with  the  centre  of 
gravity  forms  a  triangle  of  which  the  two  sides 
are  each  2,  and  as  the  altitude  is  perpendicular 
to  the  base,  the  oblique  angles  will  each  be 
45  degrees,  which  will  equal  the  required  in- 
clination. 

5.  In  this  case  the  angle  between  the  perpendicu- 
lar and  the  radius  of  the  base  will  be  30°,  and 
we  have 

tan  30°  =  \Alt.  -+  radius  of  base  ; 

Alt.  of  cone 

' 


••  —  j  •       -e^  -- 
radius  oj  base 


=  4  tan  30 


i-v/3      V3 


=  2-30940+. 

SOLUTIONS  OF  EXAMPLES. 

PAGE  154. 

1.  In  this  case  6  —  \rc  in  the  equation  of  Art.  235, 
and  sin  6  =  1  ;  hence",  by  substitution,  we  have 


2.  The  area  of  the  segment  will  equal  the  area  of 
the  sector  ACBG,  minus  the  area  of  the  trian- 
gle ACS.  Let  0  be  the  angle  ACG,  then 
ACS  =  20,  arc  AGB  =  r.26  =  2r0,  and  the 
area  of  the  sector  =  $r.2r8  =  r*8  ;  HB  —  r 
sin  6,  CH=  r  cos  0,  and  area  of  the  triangle 
ACB  =  r*  sin  8  cos  8. 


TO  ELEMENTARY  MECHANICS.  125 

Hence 


r2  (9  —  sin  9  cos  6)  x  —  r2^.^  — - —   —  r2  sin  6 

cos  0.  \r  cos  6 ; 
sin  0  —  sin  6  cos2/9 


.*.  x  = 


0  -  sin  0  cos  0 


0-8in0cos0 

3.  The  sphere  may  be  generated  by  the  revolution 

of  a  semicircle  about  a  diameter.  Area  of  the 
semicircle  =  %7tr*.  The  circumference  de- 
scribed by  the  centre  of  gravity  of  the  semi- 

circle, will  be  (Ex.  1),  f  -  •  2?r  =  \r.     Hence, 
according  to  Art.  233,  we  have 
volume  =  ^7zr2-f/'  =  |ws. 

4.  Referring  to  the  solution  of  Example  2  above, 

we  find, 

area  of  segment  =  (6  —  sin  B  cos  #)r2, 
which  multiplied  by  2^ric,  where  x  is  the  an- 
swer to  the  2d  Example,  gives  the  required  re- 
sult ;  hence  the  required  volume  is 


=  volume  of  the  sphere  x  sin8#. 

5.  Following  the  method  of  Art.  233,  and  using 
the  values  already  found,  we  have 


126  KEY  AND  SUPPLEMENT 

volume  =  1*6  x 


o  r 

X 


=  volume  of  the  sphere  x  sin 


=  area  of  great  circle  x 


If  0  =  90°  we  have  l-Tr/"8,  which  is  the  volume 

of  a  sphere,  as  it  should  be. 
The  volume  generated  as  in  Example  4  =  vol. 

in  Ex.  5  x  sin2  6. 

PAGE  158,  ART.  241.  —  In  the  first  edition  of  this  work, 
the  solution  of  this  problem  is  erroneous.  The  line 
EF  will  not  generally  be  tangent  to  the  curve  J/JV, 
and  hence  the  analysis  founded  on  that  supposition  is 
erroneous.  They  will  be  in  equilibrium  when  the 
point  g  is  vertically  under  C. 

Let  EyC  =  g),  then  the  equation  of  moments  will  be 


sn  <p  ; 
.'.W,.Eg.=  W,.gF. 
From  the  figure  we  have 


which  combined  with  the  preceding  equation  gives 
W 


also 


TO  ELEMENTARY  MECHANICS.  127 

and  as  the  three  sides  of  the  triangle  EGF  thus  be- 
come known,  the  angles  .ZTand  F  may  be  found  ;  for 
we  have  from  trigonometry 


where  s  =  $(a  +  b  +  c),  #,  6,  c,  being  the  sides  of  the 
triangle,  and  .fi'the  angle  opposite  the  side  c,  or  GF. 
Similarly, 


t  —  -  7  --  , 

6c 

where  F\§  opposite  a,  or  EG. 

In  the  triangle  .#0^  the  two  sides  GE  and  Eg,  and 
the  angle  included  by  them  becoming  known,  the 
angle  EGg  —  GgE  may  be  found  from  the  proportion 

EG  +  Eg  :  EC  -  Eg  :  :  tan  \(EGg  +  CgE)  :  tan 

\(EGg  -  GgE\ 
and  finally 

EGg  =  ±(EGg  +  CgE)  +  %(ECg  -  GgE). 


SOLUTIONS  OF  EXAMPLES. 
PAGE  169. 

1.  P  will  mo\7e  down.  Solving  for  sin  Equation  (3), 
p.  166  of  text,  we  have 

.  P  sin  B  -  TFsin  A     2 
2(P+  W) 

5347-71 

=  48-6  feet. 


128  KEY  AND  SUPPLEMENT 

2.  If  the  weights  be  in  equilibrium,  -0  =  0  and 
Eq.  (1),  p.  165,  becomes 

PsmB  -  WsmA  , 
---      -• 


or 

P  sin  B  -  IT  sin  A. 


3.  In  this  example  the  acceleration  —  \g  ; 

.••*  =  l-i^l  =  A^, 

which  in  Eq.  (3),  p.  166  of  the  text,  gives 


P  sin  45°  -  TFsin30°' 
or 

P  x  0-7071  -IW=1(P  +  TF); 
.-.  P  =  l-38  TF  +  . 

4.  From  the  conditions  of  the  problem,  and  the 
figure,  we  have 

CA  =  3  ft.,  AE  =  1  ft.,  FB  =  \  ft. ; 


The  cylinders  being  of  the  same  material, 
their  weights  will  be  as  the  squares  of  their 
radii;  hence 


Take  the  origin  of  moments  at  <?;  or  since 
gCis  vertical,  the  moment  will  be  the  same  if 
the  origin  be  anywhere  on  that  line.  The 
arm  of  W\  will  be  the  perpendicular  from  E 
to  the  line  Cg,  which  is 


TO  ELEMENTARY  MECHANICS.  129 

Eg  .  sin  EgG  =  Eg  .  sin  cp. 
Similarly,  the  arm  of  W%  will  be 

gF.  sin  cp, 
Hence  the  equation  of  moments  will  be 

Wl  .  Eg  sin  cp  =  Wz.  gFs'm  cp, 


or,  substituting  the  value  of  Wi  and  cancelling 
sin  c 


But 

£g 

substituting, 

5.^=  lift; 

.-.Eg  =  ^K. 
and 


=l  ft. 

From  trigonometry  we  have 

CEF_  (CE?  +  (EF?  -  (OF? 
WE.EF 

_  4  +  2j-  -  6i  _ 

"  ' 


and  we  have 

tan  JTCJ7  =  ^  =  ^  =  0-15; 

.'.£Cy=8°  31'  50"  +. 
5.  The   mass   moved   remains   the   same,  but  the 


130  KEY  AND  SUPPLEMENT 

effective  moving  force  is  reduced  by  the 
amount  of  the  friction ;  hence  we  have  at 
once 

P-  TFsin^  -  ^TTcos^  .    ~ 


47.-*  I/I/  ci  T\     A   —  —     it  \A/  r*r\£.    A 
i  ff      mil    ^L  iA,    f f     \s\-f !j   ^j.    f*. 

-  P+   w~  ~98_ 

P  —  TFsin  A  —  p  TTcos  A 
P+  W 

P  —  Tf(sin  A  +  n  cos  A) 

P  +  W 
Similarly, 

/r  P+W  2.n 

L-P  —  TF(sin  A  +  fit  cos  A)   g  J 


6.  Let  R  =  AC,  r  =  AE^ 

.'.  EC  =  E  -  r  =  OF,  EF=  2r. 
EF  is  bisected  by  a  perpendicular  from  (7; 


-f-  —  . 
R  —r 

By  moments,  as  in  Example  4,  we  find 


Since  Cg  is  not  perpendicular  to  EF,  the  tri- 
angle is  oblique,  and  we  have 

EG  +  Eg  :  EC  -  Eg  :  :  tan  \(ECg  +  EgC): 

ten±(ECg  -  EyC), 
or 

R  _  .ir  :  R  -  %r  :  :  tan  £(180  -  E)  :  tan  %(ECg 
-EgC}; 


TO  ELEMENTARY  MECHANICS. 

from  which  the  angle  ECg  may  be  found.  The 
angle  may  be  found  by  means  of  right-angled 
triangles.  Thus,  let  fall  a  perpendicular  from 
C  upon  EF,  and  let  the  foot  be  represented 
by  G,  which  the  reader  can  supply  in  the  fig- 
ure. Then 


CG  = 

and 

tan  ECG  =  r 


.-.  ECG  =  tan-1   /  = 

VI?  -  ZRr 

Also 


tan 


.'.</CG  =  tw-1- 


and,  finally, 

ECg  =  ECG  -  gCG. 

7.  The  velocity  of  discharge  will  be 


v  = 

and  the  quantity  discharged  in  one  second  will 
be 

k  Vfyh, 

and  in  the  time  t  the  quantity  of  discharge 
will  be 


132  KEY  AND  SUPPLEMENT 

Jet 
hence 


SOLUTIONS  OF  EXAMPLES. 

PAGE  179. 

1.  The  formula,  p.  171  of  text,  becomes 

f=  W  sin  A 


=  50 


cos  <p 

sin  30° 
cos  (60° -30°) 

i 


=  28-8  Ibs. 

2.  "We  will  have 

^=60     Bin4f°     =601bs. 
cos(— 45  ) 

3.  We  find 

p=  W^^=  Fsin^lbs. 
cosO 

4.  We  will  have 

&=  W      S1'n  A.^  =  W  tan  A  Ibs. 
cos  (—A) 


TO  ELEMENTARY  MECHANICS.  133 

5.  From  Article  254  we  have,  in  reference  to  mo- 

tion down  the  plane, 

j-  0-2  cos  (30°)          5()  _          j 
~cos  (-5°)  -0-2  sin  (-5°)  & 
up  the  plane 

F=    *  +  0*«»yr     50  =  34.41bs. 

cos  5   —  0-2  sin  5 

PAGE  180. 

6.  We  will  have,  from  the  first  value  of  F,  p.  172 

of  the  text, 


F  =          -  0.15  x  10Q  =  fio 

7.  From  Article  256  we  have 

.     A      eg 

sin  A  =  -f- 

M 

But  from  Example  4,  p.  146  of  the  text, 
where  r  =  iJR,  we  find  (see  solution  in  this 
Key) 

eg  =  £&  -  H  =  fB  ; 
/.  sin  A  -  |  ; 

.:A  =  9°  35'  40. 

8.  From  the  formula  of  Article  259  we  have 

Cc  =  H  X  4  =  3  feet, 

and,  therefore,  the  three  sides  of  the  triangle 
OOo  become  known  ;  hence,  from  trigonome- 
try, we  have 


134  KEY  AND  SUPPLEMENT 


irv        A/%s(s  — 

cos  iu=  \  -±  —  j- 
ao 


V55EE3L 

12 


=  0-96824; 
.-.  C  =  28°  57'  20." 

9.  The  formula  of  Article  259  gives 

P=  W. 

The  triangle  will  be  equilateral,  hence  each  of 
the  angles  will  be  60°. 

10.  The  equation  on  page  179  gives 

F=CD=      JL_ 


=  y3  feet. 

ANSWERS  TO  EXERCISES. 

PAGE  180. 

1.  It  will. 

2.  It  cannot  without  a  force  to  hold  it. 

3.  "When  the  centre  of  gravity  is  highest  the  poten- 

tial energy  will  be  greatest,  and  least  when 
lowest.     In  Fig.  113  it  is  greatest. 

4.  In  Fig.  114  the  potential  energy  is  a  maximum. 

In  Fig.  115  the  potential  energy  is  least. 

5.  In  Fig.  117  it  is  indifferent,  for,  since  the  weights 

are  in  equilibrium  at  all  points,  on  the  curve, 


TO  ELEMENTARY  MECHANICS.  135 

if  the  weight  W  be  moved  from  one  position 
to  another  on  the  curve,  the  weight  P  will  be 
raised  or  lowered  so  that  their  common  centre 
of  gravity  will  remain  at  the  same  height. 

The  same  general  relations  hold  in  Figs.  118 
and  119 ;  hence  the  potential  energy  is  indif- 
ferent Jn  these  also. 

PAGE  181. 

6.  In  this  case  they  are  in  equilibrium  in  only  one 

position.  If  P  >  W,  we  will  have  e  >  1,  and 
the  value  of  y  will  be  imaginary  ;  hence  W 
necessarily  exceeds  P. 

7.  They  cannot,  for  the  eccentricity  would  be  less 

than  unity. 

8.  If  the  weight  W  be  at  the  upper  extremity  of 

the  axis,  at  A,  there  will  be  equilibrium. 

9.  The  length  will  equal  twice  the  diameter  of  the 

bowl,  and  it  will  rest  on  the  edge  F. 

10.  Horizontal. 

11.  It  would  rest  on  the  horizontal  plane. 

PAGE  186. 

The  expression  for  the  moment  in  the  4th  line  of 
page  186  should  be 

F^  cos  A  -  yl  cos  a  j), 

for  reasons  given  in  Article  176  of  this  Key.     This  is 
the  same  as 

Fl  fa  sin  ^  —  yl  cos  aj). 

The  signs  of  all  the  other  expressions  on  that  page 
should  also  be  changed  from  +  to  —  and  —  to  + . 


136  KEY  AND  SUPPLEMENT 

SOLUTIONS  OF  EXAMPLES. 
PAGE  192. 

1.  t  =  0     *' „.    5  100  =  90-5  Ibs. 

2  x  0-bb33 


2.  Let 

AD  =  u,  DC=  v, 

t  =  the  tension  of  AB, 
t±  =  the  tension  of  BC', 
then 

x  +  y  =  10,  (1) 

u  +  v  =  5,  (2) 

t  =  2*!  ;  (3) 

and   the  last  formula  of  Article   269  of   the 
text  gives 
sec  BAD  =  2  sec  BCD, 

or 

-  =  2^.  (4) 

u         v 

From  the  figure, 

~  y  \  / 

Eliminating   between  equations  (1),    (2),  and 
(4)  gives 

"=»=«•  (6> 

Equations  (1),  (2),  and  (5)  give 


TO  ELEMENTARY  MECHANICS.  137 

a?  -  u2  =  (10  -  a;)2  -  (5  -  u}\ 
or 

4x  -  2w  =  15.  (7) 

Combining  (6)  and  (7)  gives 

x  =  447  +  ft., 
u  =1-44  +  ft; 
.-.  y  —  5-53  +  ft, 
v  =  3-56  +  ft 

PAGE  193. 

3.  These  conditions  require  that  the  points  A  and 

C  shall  not  be  in  the  same  horizontal  ;  and  in 
the  values  of  t  and  #t,  page  188  of  the  text, 
BAD  =  0,  and  we  have 

t=  W  cot  BCD, 

t1  =  WcosecBCD 

W 

"  sin  BCD  ' 

4.  The  angle  DAC  will  be  45°,  and  the  last  formu- 

las on  page  191  of  the  text  give 

c  cos  45°  =  t 

c  sin  45°  =  250  Ibs.  ; 


=  250  Ibs. 


138  KEY  AND  SUPPLEMENT 

5.  The  last  equation  on  page  191  of  the  text  gives 

TFsin  CAD  =  %  TF; 

.-.  sin  CAD  =  £ ; 
hence 

CAD  =  30°, 

and  the  depth  CD  will  be  one-half  the  length 
of  the  rafter. 

6.  The  equation 

£  =  |TFcot  CAD, 
page  191  of  the  text,  gives 

W=$Wcoi  CAD-, 

.'.  cot  CAD  =  2  ; 

.-.  CAD  =  26°  33'  54". 

ANSWERS   TO    EXERCISES. 

PAGE  193. 

1.  They  will  not.  The  two  added  forces  R  and  —  R 

will  form  a  couple. 

2.  They  will  not,  for  the  resultant  of  F  and  f\ 

combined  with  —  R  will  constitute  a  couple. 
If  the  force  equal  to  R  were  to  act  in  a  direc- 
tion opposite  to  the  resultant,  then,  in  Exercise 
1,  there  will  be  a  resultant  equal  and  oppo- 
site to  the  fourth  force,  and  in  the  2d  Exercise, 
after  the  3d  force  be  removed,  there  will  be  a 
resultant  equal  to  27?. 

3.  No  modification  is  needed,  for  the  moment  of 


TO  ELEMENTARY  MECHANICS.  139 

the  components  will  equal  tlie  moment  of  the 
single  force. 

4.  It  cannot — to  do  so  would  require  an  infinite 

tension. 

5.  If  the  weight  at  B  is  free  to  adjust  itself,  the 

tension  on  each  will  be  equal,  and  each  equal 
to£TF. 

6.  The  tension  will  remain  the  same.     The  tension 

is  dependent  only  upon  the  weight  and  slope 
of  the  parts,  as  shown  by  the  equations  on 
p.  188  of  the  text, 

7.  An  ellipse  for  the  sum  of  the  distances  from  the 

fixed  points  A  and  C  will  constantly  equal  the 
length  of  the  string. 

8.  Decreased — and  the  thrust   at  the  lower   ends 

will  be  diminished.  See  the  last  equations  of 
Art.  271  in  the  text. 

9.  The  thrust  and  stresses  on  the  braces  will  both 

be  increased. 

10.  If  the  strut  supports  the  weight,  there  will  be 
no  stress  on  the  rafters. 

PAGE  197. 

Galileo  was  the  first  writer,  of  whom  we  have  any 
knowledge,  who  established  formulas  for  the  strength 
of  beams.  His  work  was  published  at  Bologna  in 
1656.  Although  the  hypotheses  upon  which  the  for- 
mulas were  founded  were  false,  yet  the  law  of  variation 
of  strength  which  he  deduced  for  rectangular  beams 
was  correct.  This  law  is — the  strength  varies  directly 


140  KEY  AND  SUPPLEMENT 

as  the  first  power  of  the  breadth  and  the  square  of 
the  depth  jointly,  and  inversely  as  the  first  power  of 
the  length  of  the  beam.  But  the  factor  used  by  him 
for  determining  the  value  of  the  strength  was  three 
times  too  large. 


SOLUTIONS  OF  EXAMPLES. 

PAGE  201. 

1.  Let  w  =  the  load   per   unit    of    length  of  the 
beam,  I  =  the  length  of  the  beam,  then  will 

W=  wl 

We  have  from  the  last  equation  on  page  200 
of  the  text, 


*K<P 

500  x  8  x  S  x  12 
^  *     1400  x  8  x  8 

=  3T3j  inches. 

REMARK. — It  is  best  to  reduce  all  the  dimensions 
to  inches,  since  the  tabular  value  of  R  is  given 
per  square  inch. 

2.  From  equation  (2)  of  Art.  282,  we  have 

bd?  = 
But 


TO  ELEMENTARY  MECHANICS. 


_  .,  1000  x  8  x  12 
1200 

=  30; 

.-.  ~b  =  3-10  +  inches; 
and 

d  =  12-43  +  inches. 

3.  From  Problem  3,  p.  200,  we  have 

/7>  -  s  — 
=  *  ltb 

-  3    8000  x  12  x  12 

2  x  10000 
=  86-4; 
.-.  d  =  9-29  inches. 


4.  From  the  same  equation  as  the  preceding,  we 
have 


_  3  20000  x  10  x  12 
4x9x9 

=  11111-1  +  Ibs. 

5.  The  required  stress  will  be  the  value  of  J2  found 
from  the  equation  above, 


142  KEY  AND  SUPPLEMENT 


20000  x  lj  x  (3j)2 
10000 

=  24-5  inches. 


6.  Problem  4,  p.  200  of  the  text,  gives 


I 

12000  x  6  x  144 
15  x  12 

=  76,800  Ibs. 

7.  The  load  will  be  uniform,  and  will  equal  the 
weight  of  the  beam.     We  have 

W=2  x  2  x  I  x  4- 


and  the  formula  of  problem  2,  p.  199  of  the 
text,  gives 


W 

30000  x  2  x  4  . 

/.P  =  80,000; 
and 

Z  =  282-8  inches 

=  23  feet  6-8  inches. 


TO  ELEMENTARY  MECHANICS.  143 

PAGE  205. 

The  straight  line  of  quickest  descent  is  not  the  line 
of  quickest  descent.  Curves  of  quickest  descent  are 
called  Brachistochrones.  Their  form  depends  upon 
the  conditions  assumed.  The  forces  may  be  assumed 
to  vary  according  to  the  inverse  squares,  or  directly 
as  the  distance,  or  inversely  as  the  distance,  or  ac- 
cording to  some  other  law,  and  they  may  be  assumed 
to  act  in  parallel  lines  or  radiate  from  a  point.  If 
they  are  constant  and  parallel,  as  in  the  case  of  terres- 
trial gravitation,  the  curve  will  be  a  cycloid.  It  was 
problems  of  this  character  that  gave  rise  to  the  Cal- 
culus of  Variations — a  very  high  order  of  analysis. 

SOLUTIONS  OF  EXAMPLES. 

PAGE  208. 

1.  From  the  3d  equation,  p.  203  of  the  text,  we 
have 

2* 


200 


~  32*  x  25 
=  •2487; 
.-.  <p  =  14°  24'. 

PAGE  209. 

2.  From  the  2d  equation  on  p.  204  of  the  text  we 
have 

s  =  v0t-  \g$  sin   > 
/.  *  =  50*  -  loY2  x 
=  50*  - 


144  KEY  AND  SUPPLEMENT 

Hence,  in  3  seconds 

s  =  150  -  72|\/2  =  47-65  feet. 

At  the  end  of  5  seconds 

s  =  250  -  201&V2  =  -  34-3  ft., 

that  is,  it  will  have  ceased  to  ascend  the  plane, 

and  returning,  will,  at  the  end  of  5  seconds, 

be  34-3  feet  below  the  starting-point. 

At  the  end  of  10  seconds 

s  =  500  -  804|V2"=  -  637  3  feet  below  the 

starting-point. 

3.  The  required  velocity  will  be  the  same  as  that 
acquired  by  the  body  in  sliding  down  half  the 
length  of  the  plane  ;  hence  the  required  veloc- 
ity will  be 


v  =       -s  sn  cp 


x  100  x  T2/<r 
=  25-36  feet. 

If  a  body  starts  from  the  middle  of  the  plane 
at  the  same  time  as  the  one  at  the  upper  end, 
it  will  reach  the  foot  in  the  same  time  that  the 
upper  one  reaches  the  middle  ;  hence,  if  it  be 
projected  upward  with  the  velocity  acquired, 
at  the  instant  the  body  starts  from  the  upper 
end,  they  will  meet  at  the  middle  of  the  plane. 
4r.  The  point  must  be  higher  than  the  lower  ex- 
tremity of  the  diameter — otherwise  the  solu- 
tion is  not  possible.  The  required  line  will  be 


TO  ELEMENTARY  MECHANICS.  145 

the  distance  from  the  point  to  where  the  line 
cuts  the  circle  when  drawn  to  the  extremity 
of  the  diameter. 

5.  From  equation  (2),  p.  208  of  the  text, 


=  i  A/22- 4  x  5280 

=  38-2  +  feet  per  second 

=  26-05  +  miles  per  hour. 

(To  reduce  feet  per  second  to  miles  per  hour, 
multiply  the  former  by  6^  =  ||.) 

6.  From  the  5th  equation  we  have 

s  =  4-6V2, 

and  from  the  2d  (writing  s'  for  s  so  as  to  dis- 
tinguish it  from  the  preceding  s) 

v2  =  -A(/i  -  17-6X 
=  A  x  324  x  2640 ; 
4-6  x  324  x  2640 

-ir 

=  4:857-6  feet. 

7.  The  time  down  the  plane  may  be  deduced  from 

the  3d  equation  of  the  text.     We  have 


.  _  me      328,9 
i>  — 


h  -  17-6 

328  x  2640 
324 


146  KEY  AND  SUPPLEMENT 

PAGE  209. 

-  1634  sec. 
=  2-72  minutes. 

For  this  time  on  the  horizontal,  we  have  from 
the  4th  and  5th  equations 


4858 

4-6 

=  303-3  seconds, 
=  5-05  4-  minutes. 

8.  Equation  (5)  will  give  the  velocity  which  it 
must  acquire  in  moving  down  the  plane.  We 
have 

=  4~T 
_1000 

~  4-6 
=  217-3  feet  per  second. 

The  required  height  will  be  given  by  equation 
(2) ;  we  have 


h  = 


81  x  217-3  +  17-6  x  1200 

1200 
=  32-27  feet. 


TO  ELEMENTARY  MECHANICS. 

9.  Equation  (5),  page  208  of  the  text,  gives 


=   13-19  ft.  per  sec. 
Equation  (2)  gives 
Sis' 


"  h  -  17-6 

_  81  x  173-91 

7-4 
=  1903-64  ft 

The  author  once  had  occasion  to  use  the  prin» 
ciples  of  the  last  example  in  constructing  the 
approach  to  an  ore  dock  at  Marquette,  Mich. 

soLrnoxs  OF  EXAMPLES. 

PAGE  217. 

According  to  Article  299,  the  range  will  be 

2A  sin  2<r, 

where 

.J 


hence  the  range  will  be 
v 
ff 


—  sin  2« 


148  KEY  AND  SUPPLEMENT 


=  ^  sin  90° 

g 

=  25? 

=  804£  feet. 

The  greatest  height  will  be,  (Art.  301) 

h  sin2  a 
=  ^sin245° 


=  Oj   X 

=  201-04  +  feet. 

PAGE  218. 
2. 


=  2V15  x  12 
=  26-8  +  feet. 

3.  From  Article  299,  we  have 

2fc  =      x      =  t 
sin  2ar      ^  ' 

which  in  Article  300  gives 


%g  sin  a  cos  a 
or  solving  for  tan  a  gives 


tan  a  =  £— 
2a? 


sin2  a, 


TO  ELEMENTARY  MECHANICS.  149 

323  x  225 
2  x  1000 


=  3-6187 

.-.  a  =  74°  33'  9". 

PAGE  218. 

From  Article  300,  we  have 


2  sin  a 
x  15 


"2  x  -96387 
=  250-29  feet. 

From  Article  301,  we  have 
CD  =  h  sin2  a 

=  (250_p  (0.96387)> 
=  904-69  feet. 
4.  From  Article  299,  we  have 

;  * 

~ 


2  sin  2a 


,.,=*/! 


2  x  32i  x  25000 


2x1 
=  89G-8  +  feet 

From  Art.  301,  we  have 
A  =  h  sin2  a 


150  KEY  AND  SUPPLEMENT 


=  6,250  feet. 

From  Article  300,  we  have 
sin  a 


2  x  897  x 


1  = 


=.  89  +  seconds. 

5.  We  have  from  Articles  299  and  301, 

AS  =  4CD, 
or 

4A  sin  a  cos  «  =  4/i  sin2  tf, 
dividing  by  sin  a  cos  tf, 
tan  a  —  1  ; 
/.  a  -  45°. 

6.  In   equation  (3)   page   214,   make  y  =   —  150, 

which  is  negative,  because  in  Fig.  139  y  is 
positive  upwards,  and  the  point  where  it  will 
strike  the  plane,  in  this  example,  is  below  the 
point  of  starting,  tan  a  =  1,  and  we  have 

Q1     rZ 

1  ~  c\ 

~ 


2(75)*  x£' 
which  solved  for  x  gives 
x  =  271-5  feet. 


TO  ELEMENTARY  MECHANICS.  151 

7.  Iii  this  example 

a  -  -  30°,  y  =  -  25  feet. 
The  velocity  with  which  the  body  will  leave 
the  eaves  will  equal  that  of  a  body  falling 
through  the  vertical  height  of  the  ridge  above 
the  eaves,  and  this  value  will  be  considered  as 
the  velocity  of  projection.  We  then  have 

•y2  =  2#r  x  7  =  14gr, 

and  these  substituted  in  equation  (3),  page  214, 
give 

-  25=-  0.57785,  - 
which  solved  gives 

x  =  17-6  +  feet. 

8.  Substituting  the  values  given  in   the   example 

for  a  and  y  in  equation  (3)  page  214  gives 


60  =  400  tan  a  - 


50  =  600  tan  a  - 


s  a 

2 


32j(600)! 
2^  cos2  a 


Multiplying  the  first  by  9  and  the  second  by 
4,  and  subtracting  the  latter  from  the  former, 
gives 

3 10  =  1200  tan  a; 

/.  tan  a  =  -jrj, 
and 


152  KEY  AND  SUPPLEMENT 

a  =  15°  49'  9". 

Substituting  this  value  in  the  first  of  the  pre- 
ceding  equations  gives 


60  =  li  x  400 


32i(400)* 


_ 
2  x  (0-96213)2  x  «2 


=  228-2  feet  per  second. 


EXERCISES. 

PAGE  219. 

1.  Zero. 

2.  The  velocity  of  projection  being  the  same,  they 

will  strike  the  sea  at  the  same  time,  and  their 
range  from  the  point  where  the  ship  will  be  at 
that  time  will  be  the  same  ;  but  not  the  same 
if  reckoned  from  the  point  of  projection. 

3.  15  miles  per  hour  =  Hu^H-  fee*  per  second 

=  22  feet  per  second  ;  hence  the  actual  veloc- 
ity will  be  11  feet  per  second  in  the  direction 
of  motion  of  the  ship  in  reference  to  the  point 
from  which  the  projection  is  made. 

4.  In  reference  to  the  point  on  the  earth,  it  will  be 

the  same ;  but  not  in  reference  to  the  point  in 
space  from  which  the  projection  is  made. 

5.  It  will  reach  it  in  the  same  time.     A  horizontal 

motion  does  not  affect  the  time  of  descent  due 
to  gravity.  The  projectile  falls  from  the 


TO  ELEMENTAKY  MECHANICS.  153 

highest  point  of  its  path  (in  a  vacuum)  in  the 
same  time  that  it  would  fall  vertical  down- 
wards. 

6.  They  will  ;     for  according  to  Article  304  of  the 

text,  we  have  for  equal  ranges  the  angle  a  and 
90°  -  a.  Let  a  =  45°  -  6 ;  then  will  the  an- 
gles be  45°  -  <$  and  90  -  (45°  -  6)  =  45° 
+  <?;  but  45°  +  6  is  the  complement  of  45° 
-6. 

7.  The  lines  will  be  the  sides  of  an  angle,  and  since 

the  velocities  are  uniform,  they  will  be  divided 
into  equal  parts  in  equal  times,  by  the  motion 
of  the  bodies ;  hence,  by  geometry,  the  lines 
passing  these  equal  divisions  will  be  parallel. 
PAGE  223. 

The  relation  between  centripetal  and  centrifugal 
forces  has  been  the  subject  of  much  discussion.  In 
an  article  which  appeared  some  time  since  in  u  Na- 
ture" it  was  asserted  that  the  teim  centrifugal  force 
had  done  much  harm  in  mechanical  science,  and 
ought  not  to  be  used.  The  basis  of  the  trouble  with 
such  writers  is,  they  consider  that  centrifugal  force  is 
to  be  applied  to  the  same  body  as  the  centripetal ;  but, 
as  stated  in  the  text,  such  is  not  the  case.  Centripe- 
tal force  is  generally  conceived  to  be  the  action  of  the 
ruling,  or  larger  body,  upon  the  smaller  one,  while 
centrifugal  force  is  the  equal  opposite  action  upon  the 
other  body.  Thus,  if  a  nail  holds  one  end  of  a  string, 
while  a  body  attached  to  the  other  end  is  made  to  ro- 
tate rapidly  about  it,  the  nail  represents  the  ruling 
body,  and  the  centripetal  force  is  the  pull  of  the  string 
upon  the  rotating  body,  and  the  centrifugal  force  the 
7* 


154  KEY  AND  SUPPI^EMENT 

pull  of  the  string  upon  the  nail.  Similarly  the  at- 
tractive force  between  the  earth  and  sun  is  centripe- 
tal if  applied  to  the  earth,  and  centrifugal  if  applied 
to  the  sun.  These  are  illustrations  of  the  third  Law 
of  Motion — that  action  and  reaction  are  equal  but 
opposite. 

A  constant  centripetal  force,  caused  by  uniform 
motion  in  a  circle,  does  not  produce  an  acceleration, 
for  it  does  not  act  along  the  same  line.  The  action 
being  constantly  normal  to  the  path,  its  effect  is  con- 
stantly expended  in  deflecting  the  body  from  a  recti- 
lineal path.  Should  it  cease  to  act  as  soon  as  the 
deflection  is  made,  the  body  would  move  in  a  right 
line,  in  accordance  with  the  First  Law  of  Motion. 

The  analysis  for  determining  the  value  of  the  cen- 
trifugal force  in  the  text  is  lengthy  but  strictly  logical. 
The  following  solution  may  be  more  acceptable.  As- 
sume that  the  body  is  moving  around 
the  circle  at  a  uniform  rate  ;  then 
will  the  centrifugal  force  be  con- 
stant, and  at  any  point,  as  B,  the 
direction  of  motion  will  be  that  of 
the  tangent  BD.  The  centripe- 
tal force  must  be  such  as  to  draw 
the  body  from  the  tangent  BD  a 
distance  equal  to  DC  in  the  same  time  that  it  would 
move  over  BD.  Draw  EG  and  CG,  and  a  chord  BC. 
If  now  the  point  C  be  indefinitely  near  B,  the  limit 
of  the  sine  EC  (or  its  equal  BD),  and  of  the  chord 
BC  will  be  the  arc  BC. 

Let  BD  be  the  space  moved  over  in  time  t,  and  v 
the  constant  velocity,  then 


TO  ELEMENTARY  MECHANICS.  155 

*  =  ™> 

since  the  motion  in  the  arc  is  uniform  ;  and  since  the 
centrifugal  force  is  constant,  the  space  BE  will  be 
given  by  equation  (2),  p.  12  of  the  text,  or 

BE  =  $v't, 

where  v'  is  the  velocity  which  would  be  produced  by 
the  centripetal  force  in  passing  over  the  space  BE'm 
time  t,  if  the  force  acted  along  the  line  BE.  But 
since  the  times  are  equal,  we  eliminate  t,  between 
these  equations,  and  find 

1BE 


From  the  figure  we  have,  since  EC  is  a  mean  pro- 
portional between  BE  and  EG, 

(Ed}*  =  BE(BG  -  BE), 
which  ultimately  becomes 


_ 
BG  ' 

Substituting  this  value  above  gives 
,      2EC 


EC 


156  KEY  AND   SUPPLEMENT 

since  BG  =  2r.     But  ultimately  the  velocity  along 
BD  or  EC  is  that  along  the  arc  BC,  and  ultimately 


where  v  is  the  velocity  along  the  arc.     This  substitu- 
ted, gives 


v       ID 


t 

or,  multiplying  by  m, 
mv' 


But  the  left  member  is,  according  to  Article  122,  the 
value  of  a  constant  force,  hence 


the  required  result. 

PAGE  229,  ARTICLES  319,  320.  —  Sir  Isaac  Newton  con- 
ceived the  fact  that  if  the  attraction  of  gravitation 
varied  as  the  inverse  square  of  the  distance  from 
the  centre  of  the  force,  it  ought  to  account  for  the 
motion  of  the  moon  ;  that  is,  the  force  of  gravity  ex- 
erted by  the  earth  should  just  equal  that  necessary  to 
cause  the  proper  deviation  of  the  moon  from  a  tan- 
gent to  its  orbit.  His  first  efforts  to  prove  this  law 
failed,  due  to  the  fact  that  an  erroneous  value  of  ./?, 
the  radius  of  the  earth,  was  used.  Instead,  however, 
of  abandoning  the  idea,  and  attempting  to  account 
for  the  motion  according  to  any  other  hypothesis,  he 


TO  ELEMENTARY  MECHANICS.  157 

returned  to  his  calculation  from  time  to  time,  but 
with  no  better  results.  Finally,  while  attending  a  lec- 
ture in  London,  he  obtained  a  corrected  value  of  the 
radius,  which,  when  substituted  in  the  equation  he 
had  so  often  reviewed,  established  his  theory.  He 
was  so  overcome  by  the  grandeur  of  the  problem  as 
the  final  proof  was  becoming  apparent,  that  he  was 
unable  to  complete  the  numerical  reduction,  and 
called  a  friend  to  do  it  for  him. 

It  will  be  seen,  in  Article  319,  that  the  radius  of 
the  earth  enters  the  formula,  in  determining  the  dis- 
tance of  the  moon,  in  the  expression  60-367?.  The 
value  of  R  which  he  at  first  used  was  too  small  by 
iV  to  iV  °f  i*8  true  value.  See  also  remarks  on  pp. 
36  and  37  of  this  Key. 

The  law  of  gravitation  was  not,  at  once,  universally 
accepted.  Several  times,  especially  in  the  history  of 
astronomy,  certain  phenomena  appeared  to  conflict 
with  this  law,  when  it  was  called  in  question,  and  its 
truth  assailed.  But  all  opposition  to  it  disappeared 
after  Laplace,  by  his  truly  wonderful  analysis,  ex- 
plained all  those  paradoxes,  and  accounted  for  all  the 
motions  of  the  solar  system,  on  the  simple  law  of 
Universal  Gravitation.  It  is  now  believed  to  be  true, 
not  only  for  the  solar  system,  but  for  every  particle 
of  matter  in  the  universe.  Newton  believed  that  the 
ether  of  space,  whatever  it  might  be,  was  more  dense 
in  the  vicinity  of  the  planets,  than  in  remote  space ; 
that,  indeed,  it  might  be  only  air  extremely  rarefied. 

To  find  the  stress  due  to  the  attraction  between  the 
earth  and  moon. 

It  equals  the  centrifugal  force,  the  value  of  which  is 


158  KEY  AND  SUPPLEMENT 


The  mass  m  of  the  moon  is  about  3£  times  that  of  a 
mass  of  water  of  equal  volume,  and  as  a  cubic  foot  of 
water  weighs  62£  Ibs.  when  g  =  32£  feet,  and  the 
diameter  of  the  moon  is  2,160  miles,  we  have 


m  =  3i  x  ITT  (2160  x  5280)3  X         Ibs. 


The  time  of  the  revolution  of  the  moon  about  the 
earth  is  about  27£  days  ;  hence  the  angular  velocity 
per  second  is 

27T 


6?  = 


X  24  X  3600 ' 


The  mean  distance  between  the  centres  of  the  moon 
and  earth  is  about  240,000  miles  ;  hence 

r  =  240,000  X  5280  ; 

all  the  magnitudes  being  in  feet,  and  all  the  times 
reduced  to  seconds.  Hence  we  have 

_  3£  x  |7T8  x  (2160)3  X  (5280)4  x  240000  x  62j 
32£  x  (27i  X  24  x  3600)2 

which  reduced  gives,  approximately, 

44,000,000,000,000,000,000  Ibs. 
=  44  x  1018. 

A  steel  rod  one  square  inch  of  section  will  sustain  a 
pull  of  120,000  Ibs. ;  hence  it  would  require  (approx- 
imately) 


TO  ELEMENTAKY  MECHANICS.  159 

370,000,000,000,000 
=  37  X  1013  square  inches 

of  steel  to  hold  the  moon  in  her  orbit  if  substituted 
for  the  attraction  between  the  earth  and  moon. 
In  one  square  mile  are 

4,014,489,600  sq.  inches ; 

which,  divided  into  the  above,  gives,  for  the  equiva- 
lent section  in  miles, 

90,000  sq.  miles  nearly. 

Since  the  radius  of  the  moon  is  1,080  miles,  the  area 
of  a  great  circle  will  be 

3,660,000  sq.  miles  nearly. 

which,  divided  by  the  solid  section  of  the  steel  rod, 
gives  40  +  ;  hence,  if  the  rods  were  each  one  square 
inch  in  section,  and  the  great  circle  of  the  moon  be 
divided  into  inch-square  spaces,  the  rods  would  cover 
one  space  in  40. 

The  square  of  the  diameter  of  the  earth  is  nearly 
15  times  the  square  of  the  diameter  of  the  moon,  hence 
such  a  steel  rod  would  cover  about  -^  of  the  merid- 
ian circle  of  the  earth. 

If  the  material  be  iron  instead  of  steel,  and  if 
10,000  Ibs.  be  taken  to  represent  the  tenacity,  a  value 
quite  commonly  used  in  engineering  structures,  the 
rod — or  rods — would  cover  more  than  one-fourth  the 
cross-section  of  the  moon,  and  about  -fa  of  a  great  cir- 
cle of  the  earth. 

The  same  problem  applied  to  the  attraction  between 
the  sun  and  earth  gives 


163  KEY  AND  SUPPLEMENT 


x  f  7r3(20500000)3  x  5280  X  92500000  X  62j .. 
32£-  x  (365  x  24  X  3000)2 


where  it  is  assumed  that  the  mean  density  of  tho 
earth  is  5|  times  that  of  water,  the  distance  between 
the  centre  of  the  sun  and  the  earth  92,500,000  miles,  the 
radius  of  the  earth  20,500,000  feet,  and  the  time  of 
the  revolution  in  the  orbit  365  days.  This  reduced 
gives,  approximately, 

912  x  1018  Ibs., 
or 

912,000,000,000,000,000,000  Ibs., 

or  more  than  20  times  that  between  the  earth  and 
moon.  According  to  this  result  it  would  require  a 
solid  steel  rod  of  a  cross-section  equal  nearly  to  one- 
half  the  great  circle  of  the  moon,  the  tenacity  being 
120,000  Ibs.  per  square  inch  ;  or  if  the  rod  be  of  iron, 
and  10,000  Ibs.  be  used  for  its  tenacity,  the  section  of 
the  rod  will  be  about  f  of  the  area  of  a  great  circle  of 
the  earth.  These  examples  show  the  immense  stress 
of  gravitation  when  large  masses  are  involved. 

The  following  examples  will  show  that  the  same 
force,  under  certain  circumstances,  is  comparatively 
weak. 

Required  the  time  that  it  will  take  two  spheres  of 
the  same  material  as  the  earth,  each  one  foot  in  diam- 
eter, placed  12^  inches  from  centre  to  centre,  to  come 
together  ly  their  mutual  attractions,  in  void  space. 


TO  ELEMENTARY  MECHANICS.  161 

According  to  one  of  Newton's  laws  of  attraction, 
the  force  varies  as  the  mass.  If  the  diameter  of  the 
earth  be  41,700,000  feet  (p.  33  of  the  text),  then  will 
the  mass  of  the  sphere  1  foot  in  diameter  be 


and  therefore  the  acceleration  at  the  surface  of  the 
earth  due  to  the  attraction  of  such  a  sphere  placed  at 
the  centre  of  the  earth  would  be 

g 


(41700000)3  ' 

an  inappreciable  quantity. 

According  to  another  of  Newton's  laws,  combined 
with  the  analysis  on  pp.  33  and  34  of  this  Key,  the 
force  varies  inversely  as  the  square  of  the  distance 
from  the  centre  of  the  sphere  ;  hence  at  the  distance 
of  one  foot  from  the  centre  of  the  small  sphere  we 
have 

P:  (20850000)- ^-p^^^:/; 
-p  -  9 


8  x  20850000 ' 

for  tne  acceleration  of  a  particle  one  foot  from  the 
centre  of  the  sphere.  This  will  also  be  the  accelera- 
tion of  any  uniform  sphere  whose  centre  is  one  foot 
from  the  first  sphere,  if  the  diameter  of  the  second 
sphere  does  not  exceed  one  foot ;  or  should  it  exceed 
that  diameter,  it  will  still  be  true  for  the  mass  of  the 
sphere  if  reduced  to  a  sphere  whose  diameter  is  less 


162  KEY  AND  SUPPLEMENT 

PAGE  229. 

than  one  foot,  and  hence  will  be  the  acceleration  pro- 
duced upon  an  equal  sphere.  If  the  first  sphere  were 
fixed  in  space,  the  second  sphere  would  move  the  £  of 
an  inch  ;  but  as  both  are  free  each  will  move  one- 
half  the  distance  between  them,  or  i  of  an  inch. 

In  order  to  simplify  the  problem,  we  will  assume 
that  the  acceleration  is  uniform,  while  the  spheres  are 
moving  the  \  of  an  inch,  and  is  that  due  to  the  at- 
traction at  a  distance  of  12  inches  from  their  centres ; 
then  will  equation  (4),  page  12  of  the  text,  be  applica- 
ble, and  we  have 


2  x  \  x  Ty 


__  _  _ 

8  X  20850000 


=  V108031 

=  328-7  seconds,  nearly, 

or  less  than  5£  minutes. 

This  problem  is  in  "  The  System  of  the  World,"  by 
Sir  Isaac  Newton,  p.  527  of  our  copy  of  the  Principia. 
It  is  there  stated  that  "  the  attraction  of  homogeneous 
spheres  near  their  surfaces  are  (Prop.  Ixxii.)  as  their 
diameters.  Whence  a  sphere  of  one  foot  in  diameter, 
and  of  a  like  nature  to  the  earth,  would  attract  a 
small  body  placed  near  its  surface  with  a  force 
20,000,000  times  less  than  the  earth  would  do  if 
placed  near  its  surface,  but  so  small  a  force  could  pro- 
duce no  sensible  effect.  If  two  such  spheres  were 


TO  ELEMENTARY  MECHANICS.  163 

distant  but  %  of  an  inch,  they  would  not,  even  in 
spaces  void  of  resistance  come  together  by  the  force  of 
their  mutual  attraction  in  less  than  a  month's  time." 

"We  have  sought  for  the  source  of  the  error  in  the 
Principia,  by  determining  the  conditions  necessary 
for  giving  his  result,  but  have  not  satisfied  ourselves. 
We  observe  that  he  made  an  error  in  saying  that  the 
force  of  attraction  on  the  surface  of  the  small  sphere 
is  20,000,000  times  less  than  on  the  earth ;  for,  ac- 
cording to  his  proposition — considering  the  radius  of 
the  earth  as  20,000,000  feet— it  should  be  40,000,000 
times  less  ;  and  according  to  the  inverse  squares,  at 
the  distance  of  one  foot  from  the  centre  of  the  small 
sphere,  it  would  be  160,000,000  times  less.  If  now 
we  assume  that  the  particle  is  moved  i  of  afoot,  un- 
der the  action  of  this  force,  we  would  have 


=</ 


2  x 


32 


160000000 


=  V2,500,000  (nearly)  seconds, 

where  the  quantity  under  the  radical  is  nearly  the  num- 
ber of  seconds  in  one  month.  Whether  this  gives  any 
clue  to  the  source  of  the  error,  we  are  unable  to  say.* 

*  The  author  presented  the  above  result  to  the  Physical  Sec- 
tion of  the  Am.  ASBO.  for  Ad.  Sc.,  at  Montreal,  1882.  The  genu- 
ineness of  "  The  System  of  the  World,  by  Sir  Isaac  Newton,"  was 
there  called  in  question.  The  work  was  originally  issued  in  a 
separate  volume,  but  in  the  author's  copy  it  is  bound  with  the 
Principia,  and  paged  with  it.  But  it  evidently  forms  no  part  of 
the  Principia  proper. 


104  KEY  AND  SUPPLEMENT 

PAGE  229. 

The  assumptions  made  in  order  to  simplify  the 
problem,  not  being  strictly  accurate,  we  now  propose 
the  following  problem  : 

Assume  that  two  equal  spJieres  of  the  same  material 
as  the  earth,  each  one  foot  in  diameter,  are  reduced 
in  size  to  a  mere  point  at  their  centres,  and,  placed 
one  foot  from  each  other  /  required  the  time  it  would 
take  for  them  to  come  together  by  their  mutual  attrac- 
tion, they  being  uninfluenced  ly  any  external  force. 

We  first  make  a  general  solution. 
Let  E  =  the  mass  of  the  earth, 

m  =  the  mass  of  one  of  the  spheres, 
m'  =  the  mass  of  the  other  sphere, 
R  =  the  radius  of  the  earth, 
r  —  the  radius  of  one  of  the  spheres, 
r  =  the  radius  of  the  other  sphere, 
g  =  the   acceleration   due   to  gravity   on   the 

earth, 

/*  =  the  acceleration  due  to  the  attraction  of  a 
sphere   of   mass    unity   upon    another 
sphere   ot  mass  unity,  the  distance  be- 
tween their  centres  being  unity, 
a  —  the  original  distance  between  the  centres 

of  m  and  m',  and 
x  —  the  distance  between  their  centres  at  the 

end  of  time  t. 

The  values  of  units  are  determined  by  measure- 
ments on  the  surface  of  the  earth,  and  the  unit  mass 
will  be  so  taken  as  to  correspond  with  the  units  in 


TO  ELEMENTARY  MECHANICS.  165 

use.  The  acceleration  produced  by  a  mass  E,  con- 
ceived to  be  reduced  to  a  point,  upon  one  of  the  units 
of  mass  at  a  distance  unity,  will  be  E  times  as  great 
as  that  produced  by  the  other  unit,  or 


and  at  a  distance  R  it  will  be 

E 
*J?> 

which,  according  to  the  notation,  will  be  the  accelera- 
tion on  the  surface  of  the  earth  due  to  gravity  ;  hence 

•E  /i  \ 

ff  =  t*jp,  (1) 

and 

*  =  |^  (2) 

by  means  of  which  the  numerical  value  of  the  unit  of 
acceleration  may  be  determined. 

Again,  the  acceleration  produced  by  the  attraction 
of  the  mass  m  upon  one  of  the  units  of  mass  at  dis- 
tance, unity  will  be 


and  at  a  distance,  x,  the  acceleration  will  be 


166  KEY  AND  SUPPLEMENT 

winch  will  also  be  the  acceleration  produced  upon  m, 
by  the  attraction  between  m  and  m'  at  the  distance  x 
between  them,  since  the  result  will  be  the  same  as  if 
both  masses  were  concentrated  at  their  centres  of 
gravity  ;  for  all  of  m  will  exert  the  same  force  upon 
each  particle  of  m'  as  upon  each  particle  of  the  unit. 
But  the  pull  in  pounds  will  equal  the  mass  into  the 
acceleration  (p.  44,  Art.  86  of  the  text),  or 


mm 


Similarly,  considering  the  attraction  of  m'  upon  m, 
the  acceleration  produced  upon  m'  will  be 


eP 

and  the  pull  in  pounds  will  be  m  times  this  amount, 

or 

m'm  ,Cv 

^  IT '  ® 

which  is  the  same  as  (4),  as  it  should  be.  Substitut- 
ing /*  from  (2)  in  (4)  or  (6)  gives  for  the  pull  in 
pounds  (or  their  equivalent),  of  any  two  masses  m  and 
m', 

I    7~)9.  -  -I 

3-  CO 


The  origin  of  the  axis  of  x  being  at  the  centre  of 
one  of  the  masses  and  moving  with  it,  and  the  total 
mass  moved  by  the  stress  being  m  +  m',  we  have 


TO  ELEMENTARY  MECHANICS.  167 

JPx  mm'JPy     1  , 

(rn  +  m}-^--          -5p.-r,  (8) 

which  integrated   (Analyt,  Mech.,  pp.  33,  34),  observ- 
ing that  for  t  =  0,  x  =  0,  and  v  =  0,  and  that  //  in 

,       mm'I&q     .     .,. 

the  reference  equals  7 —      — £-*«  in  this  case,  gives 
(in  +  m}E 

[(m  +  m'}Ea~\±      T, 

<= L  a^'A  J x  L^-^ 

which  for  the  limits  gives 


If  both  spheres  are  of  the  same  density,  their  masses 
will  be  as  the  cubes  of  their  radii  ;  or 

r3 


and  we  have 


and  if  the  spheres  are  equal,  as  in  the  problem,  we 
have 


If  a  =  1  foot,  ^  =  20,850,000,  r  =  ±  foot,  ^  = 
we  have 


168  KEY  AND  SUPPLEMENT 

i 


=  3577  seconds,  nearly 

=  59-6  minutes,  nearly. 
PAGE  229. 

An  exact  solution  of  the  former  problem  may  be 
made  by  means  of  equation  (9),  by  substituting  in  it 
a  =  12i  inches  =  1A  feet,  and  making  x  =  a  for  one 
limit  and  1  foot  for  the  other.  We  would  thus  have 


=  384  seconds  nearly, 
or  less  than  6£  minutes. 

The  following  is  the  reduction  of  the  preceding  ex- 
pression.    To  find  the  value  of  ^"'(TOI)  '  we 

log.  12  =.  1-079181 
log.  12-25  =  1-088136 


Dif.  =  1-991045 

Dividing  by  2,  1-995523, 

and          '     log.  0-98974  -  f-995522, 
or         log.  cos  8°  12'  46"  =  9-995523. 


The  length  of  arc  will  be 

8°  12'  46" 


180°  X  60  X  60 
or 


31416, 


TO  ELEMENTARY  MECHANICS.  169 

29566 


648000 
which  may  be  reduced  as  follows  : 

log.  29566  =  4470892 

log.  3-1416  =  0497150 

ar.  co.  log.  648000  =  4-188425 


subtracting  10,  log.  0-14337  =  1-156467 
adding  log.  1-A,  log.  1-02083  =  0-008951 


gives  log.  0-14636  =  1-165418  ; 

hence, 

U-  cos-V--V  =  0-14636. 


We  also  have  (1&  -  1)  1  =  &  =  0-02083  +  which 
added  to  the  preceding  result  gives  0-16719  +  for 
the  value  in  the  brackets. 

For  the  first  parenthesis,  we  have 

log.  1A-  =  0-008951 

log.  20850000  =  7-319106 

log.  8  =  0-903090 

log.  32|,  ar.  co.  =  8-492594 


subtracting  10  and  dividing  by  2,        6-723741 


log.  2300-7  =  3-361870. 
Adding  log.  0-16719  =  1-223209 


gives  log.  384-6  =  2-585079  ; 


170  KEY  AND  SUPPLEMENT 

that  is.  the  time  will  be  385  seconds  nearly,  or  less 
than  6|  minutes. 

To  find  the  stress  in  pounds  which  would  be  ex- 
erted by  the  mutual  action  of  two  such  spheres  at  a 
distance  of  one  foot  between  their  centres,  we  have 
from  equations  (7)  and  (10),  since  m  =  m',  and  x  =  1, 

E 


The  mass  of  the  earth  is  5£  times  an  equal  mass  of 
water.  The  weight  of  a  cubic  foot  of  water  is  62£  Ibs. 
at  the  place  where  y  =  32|,  and  the  volume  of  the 
earth  is  7tR?  hence 


E  =  5|  x  C2i  X 
which  substituted  above  gives  for  the  stress 

%7t  x  5£  x  62|    „ 
20850000  x  32f        ' 
=  0-00000215  +  Ibs., 

or  less  than  joirinnnjo-  °f  a  pound,  a  quantity  inappre- 
ciably small. 

A  would-be  inventor  once  proposed  to  weigh  the 
varying  force  of  gravity  by  means  of  very  delicate 
scales,  and  by  using  them  on  board  a  steamer,  thus 
determine  whether  the  water  underneath  were  deep 
or  shallow.  Since  the  density  of  the  solid  earth  ex- 
ceeds that  of  water,  the  force  of  gravity  at  the  sur- 
face of  deep  water  will  be  less  than  on  shallow  water, 
but  it  is  evident  that  the  rocking  and  heaving  of  the 
vessel  would  probably  produce  more  effect  upon  such 


TO  ELEMENTARY  MECHANICS.  171 

delicate  mechanism  than  that  due  to  the  variations  of 
the  force  of  gravity. 

It  is  also  stated  that  mariners  have  observed  that 
two  ships  at  rest  in  a  quiet  sea  tend  to  approach  each 
other,  but  it  will  be  found  that  the  gravitating  stress 
due  to  their  mutual  attractions  is  so  small  that  it 
might  be  more  than  neutralized  by  a  very  slight 
breeze,  or  by  the  beating  of  very  small  waves. 

To  give  some  idea  of  the  magnitude  of  this  stress, 
assume  that  the  vessels  are  of  equal  mass,  and  each 
equivalent  to  a  sphere  of  the  average  mass  of  the 
earth,  each  30  feet  in  diameter,  and  200  feet  between 
their  centres. 

Conceive  that  the  masses  are  reduced  to  their  cen- 
tres, then  since  the  mutual  attraction  of  unit-spheres 
at  distance  unity  between  them  is  y-g^f  f,nro  °f  a  pound, 
the  attraction  of  the  masses  of  the  ships,  reduced  to 
their  centres,  at  the  same  distance  (one  foot)  will  be 


TOrttmnr  X  3°3  X  3('3  =  14?1  Pound8  i 
and  at  the  distance  of  200  feet,  it  will  be 

1471 


2002 


=  0.04  pound  nearly. 


If  the  distance  between  them  be  500  feet,  the  stress 
would  be 

1471 

5QQ2  =  T}TF  of  a  pound,  nearly. 

If  all  external  forces,  such  as  the  wind  and  action  of 
the  sea  were  neutralized,  this  slight  stress  would  be 


172  KEY  AND  SUPPLEMENT 

sufficient  to  cause  the  ships  in  question  to  collide  in 
a  short  time. 


SOLUTIONS  OF  EXAMPLES. 

PAGE  233. 

1.  We  have  from  equation  (5),  page  226, 

v*       W  v2 
f  =  m  -  —  —  •  —  . 
r        g     r 

equal  2  W; 

W    ?;* 

...2TF=-  •-; 

g     r 


or  the  velocity  must  be  that  acquired  by  a 
body  falling  freely  through  a  distance  equal  to 
the  radius  of  the  circle  (see  eq.  (3),  p.  36  of 
the  text). 

2.  If  the  tension  is  3  W,  we  have 

W   -u2 
3W=  -•-; 

g    >' 


Let  n  be  the  number  of  revolutions,  then  the 
velocity  will  be  the  space  (2?mi)  divided  by 
the  time,  or  60  seconds,  r  must  be  reduced  to 
feet  and  the  time  to  seconds,  for  g  is  given  in 
feet  per  second.  Hence  we  have 


TO  ELEMENTARY  MECHANICS.  173 

ftrn 


v  = 


60  30    ' 


r 
30 


3-1416 
=  38-3. 


3.  The   centrifugal  force  will   equal   the  weight; 
hence 

tf       W  v> 
W  =  m  -  =  —  •  -  ; 
r         a     r 


Ttrn   .  ,.  ,  x 

.*.  v  =  *jgr  =  -K7T-  (see  preceding  example) ; 


30 

•    <n  — 

~ 


_ 

34416      r 
=  38-3, 

as  in  the  preceding  example. 

4.  "When  the  body  is   at   the   lowest   position    its 
weight  will  be  added  to  the  centrifugal  force, 


174  KEY  AND  SUPPLEMENT 

but  the  tension  due  to  the  centrifugal  force 
equals  the  weight;  hence  the  tension  will 
equal  2  W. 

5.  The  friction  will  be  >u  times  the  pressure  due  to 
the  centrifugal  force,  and  must  equal  the 
weight.  Let  n  be  the  number  of  revolutions 
per  minute  ;  then  will  the  angular  velocity  per 
second  be 

.  n 


60/>          60 
and  the  centrifugal  force  will  be  (Art.  314) 


and  the  friction  will  be 


*•  ^WfT     ¥' 

_30,/  9 


6.  The  body  is  assumed  to  be  in  a  radial  groove, 
and  the  string  slightly  elastic  so  as  to  allow  the 
body  to  move  slightly  along  the  groove,  and 
thus  give  the  friction  a  chance  to  act. 
The  angular  velocity  per  minute  will  be 

2*  x  250 : 
and  the  centrifugal  force  will  be 


TO  ELEMENTARY  MECHANICS.  175 

f  -    "07500  *\  2     30 

7  :  7  V  eo  J  '  12 

=  iVjX  *P  (3-141  6)2?F. 

The  friction  will  be  0-15  of  this  amount,  and 
the  tension  of  the  string  0-85  of  the  same; 
hence  the  tension  will  be 

T=  0-85  x  TVv  X  *| 
=  45-27  +  pounds. 

PAGE  234. 

7.  The  friction  will  be 


The  centrifugal  force  will  be  (Art.  313), 


gR 
or 


324-  -  2500 


10 

=  89-675  feet  per  second 
=  61-14  miles  per  hour. 


17G  KEY  AND  SUPPLEMENT 

8.  According  to  Article  322,  we  have 

,       v*   I 
n  =  —  — 

/3Q  X  5280  \8    ' 
_  \  GO  X  60    / 
32£  X  3000 

—  0-09$  feet 
=  1-12  inches. 


9.  The  weight  will  be  to  the  centrifugal  force  as 
the  length  of  the  stri 
ment  of  the  body  ;  or 


the  length  of  the  string  is  to  the  lateral  raove- 


—    _ 
W 

To  find/"  we  have  equation  (5),  Art.  313, 

/40  X  5280 
f_  W\  60  X  60 

•/ 


g          4000 
which  substituted  above,  gives 

_6 /40  x  5280X8. 

X  4000  V     3600      /  ' 


x  — 


=  0460  ft. 
=.  1-92  inches. 


TO  ELEMENTARY  MECHANICS.  177 

The  value  is  independent  of  the  weight. 
10.  To  find  the  time  of  making  one  revolution, 
we  have 

T  =  60  -T-  100  =  f. 

Then  from  p.  232  of  the  text,  we  have 

gT* 

cos  cp  =  -j-f  —  j^ 
2  .  AB 

X      )8 


~  5  x  (3-1416)2 
=  0-23466  ; 

.-.  tp  =  76°  25'  43". 

Or  by  logarithms 

log.  32£  =  1.507406 
log.  (0.6)s  =  T.556303 


adding,  1 063709. 

log.  5  =  0.698970 
log.  (3.1416)2  =  0.994300 


adding,  1.693270  which  subtract- 

ed from  the  above  gives 

log.  cos  q>  =  1370439; 
.:<p  =  76°  25'  43". 

If  cos  <p  —  1,  or  q>  =  0,  a  relation  will  be  es- 
tablished between  AB  and  T,  or  we  will  have 


and  if  A  B  is  assumed  to  be  less  than  the  value 
8* 


178  KEY  AND  SUPPLEMENT 

found  by  this  formula,  cos  cp  will  exceed  unity, 
and  hence  q>  will  be  imaginary,  or,  in  other 
words,  the  conditions  will  be  impossible.  We 
also  have 

AC  =  AS  cos  <p 

=  15  cos  76°  25' 43" 

=  15  x  0-23466 

=  3-52  inches; 
and 

BC  =  AB  sin  <p 

=  15  X  sin  76°  25'  43" 

=  14-58  inches. 

11.  Let  n  —  the  number  of  revolutions  per  minute, 
=  n  -7-  60  per  second.  The  distance  BC  will  be 

BC  =14  sin  10°  inches, 
and  the  veloc  ty  in  feet  per  second  will  be 


X  44  sin  10° 

V  ^  — 

60 
=  TloTTft  sin  10°. 

Employing  the  equation  at  the  bottom  of  page 
232  of  the  text,  making  F  =  4  Ibs.,  and  sub- 
stituting the  value  of  «  given  above,  we  have 

sin  10°)2  =  32£  x  tt  X  sin  10°  tan  10°  x 

X  14  sin  10°  . 
5  x  12  ' 


TO  ELEMENTARY  MECHANICS. 


179 


180  4/193     14  F      F~  ~T~  ~T 

•     in   --   __    'I/    __  V    _         __  1  __ 

"  lit         63  U  cos  10°  ^5  sin  10°  J 


=  118-86. 

The  reduction  is  as  follows  : 


180 


7  X  31416  r      9     \4  x  0-98481  ^  5  x 


1          \ 

0'17365  / 


180 


log.  1-40557       0-147852 

log.  1351       3-130655 

log.  9  ar.  co.       9 '045757 


Dividing  by  3, 


2-324264 


1162132 

log.  180       2-255273 
log.  21 -9912  a.  c.       8  "657750 


log.  118-86       2-075055 

12.  The  tenacity  to  be  overcome  will  be  that  in  a 
section  through  the  axis  of  the  stone,  and 
hence  will  be 

4  x  12  x  4  x  600  =  115200  Ibs. 

The  centrifugal  force  producing  rupture  will 
be  that  due  to  either  half  of  the  stone  into 
which  it  is  divided  by  the  plane  section  ;  and 
this  will  equal  the  mass  of  one-half  multiplied 


180  KEY  AND  SUPPLEMENT 

by  the  distance  of  its  centre  of  gravity  from 
the  axis  of  rotation. 

The  centre  of  gravity  of  a  semicircle  is  f  - 

71 

from  the  centre  of  the  circle  (Ex.  1,  p.  154  of 
the  text). 

If  d  be  the  weight  of  a  cubic  foot  of  the 
stone,  the  weight  of  one-half  will  be 


X  thickness  X 

=  £«•  x  4  x  j-  x  <y 


and  mass 


If  n  be  the  number  of  revolutions  per  min- 
ute, then  the  angular  velocity  per  second  will 
be 

n       mt 

*60  =  3<r; 

and  the  centrifugal  force  will  be 


.-.  n  =  \/ 


115200  x  9  X  302  x  193 
8  X  2  X  (3-U16)2  x  6  X  $ 

13786-7 


TO  ELEMENTARY  MECHANICS.  181 

EXEECI8ES. 

PAGE  235. 

1.  It  will  not.     The  deviating  force  will   be    the 

normal  component  of  the  force  at  the  centre. 

2.  Because  the  centrifugal  force  causes  a  pressure 

against  the  side  of  the  vessel,  to  balance  which 
requires  an  increase  of  height  of  the  water. 

The  same  tendency  exists  with  other  sub- 
stances, but  in  order  that  any  substance  shall 
actually  be  elevated  at  the  outer  surface,  the 
centrifugal  force  must  overcome  the  friction 
between  its  particles. 

3.  The   diameter   of  the   earth   at   the  equator  is 

about  26  miles  more  than  at  the  poles.  If  the 
earth  were  changed  to  a  sphere,  the  polar 
diameter  would  be  increased  about  13  miles, 
and  the  equatorial  decreased  about  the  same 
amount.  The  exact  change  in  the  dimensions 
involves  a  knowledge  of  the  volume  of  ellip. 
soids. 

4:.  It  does.  The  angular  velocity  of  the  earth 
being  constant,  the  centrifugal  force  varies  as 
the  distance  from  the  centre. 

5.  It  would  come  to  rest  on  the  surface  of  the  hol- 

low. 

6.  According  to  Article  318,  if  the  earth  revolved 

in  84  m.  42T6T  sec.,  bodies  on  the  equator  would 
weigh  nothing ;  hence  if  it  revolved  in  84 
minutes  they  would  fly  off  if  free,  but  if  held 


182  KEY  AND  SUPPLEMENT 

by  cohesion  or  otherwise,  the  holding  force 
must  be  overcome  before  they  could  fly  off. 

7.  It  would  be  nearer. 

8.  Because  the  velocity  in  their  orbits  is  necessarily 

greater  in  order  to  balance  the  attractive  force 
of  the  sun,  and  the  circumference  of  the  orbit 
is  less  than  those  more  remote.  Kepler's  law- 
is — the  squares  of  the  times  of  the  revolutions 
are  as  the  cubes  of  the  mean  distances  from  the 
sun. 

9.  The  centrifugal  force  is  not  destroyed — it  is  only 

equilibrated. 

10.  Tangentially. 

11.  It  is  proper  to  say  that  the  centrifugal  force — 

and  hence  the  centripetal — is  due  to  the  veloc- 
ity of  the  stone  in  the  sling.  Strictly  speak- 
ing, the  centripetal  force  has  nothing  to  do 
with  the  velocity ;  but  indirectly  it  has,  for  the 
velocity  could  not  be  produced  without  the 
centripetal  force — and  it  is  only  in  this  sense 
that  it  has  something  to  do  with  it. 

12.  The    centrifugal    force    causes    the   clothes   to 

press  against  the  vertical  inside  surface  of  the 

vessel. 

13.  They  are  not  affected  by  the  rotation  ;  but  if 

the  earth  should  cease  to  rotate,  the  surface 
at  the  poles  would  be  elevated ;  and  hence 
cause  bodies  there  to  weigh  less  by  being  at  a 
greater  distance  from  the  centre  of  the  earth. 


TO  ELEMENTAKY  MECHANICS.  .       183 

14.  There  is.  If  the  moment  of  the  centrifugal 
force  in  reference  to  the  outer  rail  as  an  ori- 
gin exceeds  the  moment  of  the  weight  of  the 
car  in  reference  to  the  same  point,  they  will 
overturn. 

PAGE  236. — The  analysis  of  this  chapter  very  properly 
belongs  to  higher  analysis ;  but  we  have  succeeded, 
by  means  of  curves  and  special  artifices,  in  bringing  it 
within  geometrical  and  algebraic  analysis. 

In  the  language  of  the  calculus,  if  m  be  the  mass  and  rj  the 
force  at  a  unit's  distance,  we  would  have 
cPx_ 

Multiplying  by  dx  gives 

dx  d?x 
m        y    =  —  773;  dx, 

and  integrating, 


dx'1 


Assuming  for  the  initial  conditions  that  the  velocity  is  zero,  and 
x  =  a,  we  have 

0  -  -  yd?  +  C, 
,-.  <7=7<»« 
and  we  have 


which  is  the  square  of  the  velocity,  and  is  the  same  as  equation 
(2),  page  233  of  the  text.     From  the  last  equation  we  have 


which  integrated  gives 


184  KEY  AND  SUPPLEMENT 

fm          .  x        ™ 
t  =  {/  -  sin"1  — h  C . 
'    T)  a 

Assuming  that  t  =  0  for  x  =  a,  we  have 


TO 
— 

77 
If  a;  =  0,  we  have 


But  we  also  have  sin—1  0  =  ?r,  for  which  value,  we  have 


which  is  the  same  as  Equation  (4),  page  241  of  the  text. 

PAGE  243,  ART.  328. — Captain  Kater  used  the  principle 
of  the  convertibility  of  the  centres  of  suspension  and 
oscillation  for  determining  the  length  of  a  simple 
seconds  pendulum,  and  hence  the  acceleration  due  to 
gravity.— Phil.  Trans.,  1818. 

Let  a  body,  furnished  with  a  movable  weight,  be 
provided  with  a  point  of  suspension  A,  and  another 
point  on  which  it  may  vibrate,  fixed  as  nearly  as  can 
be  estimated  in  the  centre  of  oscillation  B,  and  in  a 
line  with  the  point  of  suspension  and  the  centre  of 
gravity.  The  oscillations  of  the  body  must  now  be 
observed  when  suspended  from  A,  and  also  when  sus- 
pended from  B.  If  the  vibrations  in  each  position 
should  not  be  equal  in  equal  times,  they  may  readily 


TO  ELEMENTARY  MECHANICS.  185 

be  made  so  by  shifting  the  movable  weight.  When 
this  is  done,  the  distance  between  the  two  points  A 
and  B  is  the  length  of  the  simple  equivalent  pendu- 
lum. Thus  the  length  L  and  the  corresponding  time 
T  of  vibration  will  be  found  uninfluenced  by  any 
irregularity  of  density  or  figure.  In  these  experi- 
ments, after  numerous  trials  of  spheres,  etc.,  knife 
edges  were  preferred  as  a  means  of  support.  At  the 
centres  of  suspension  and  oscillation  there  were  two 
triangular  apertures  to  admit  the  knife  edges  on 
which  the  body  rested  while  making  its  oscillations. 

Having  thus  the  means  of  measuring  the  length  L 
with  accuracy,  it  remains  to  determine  the  time  T. 
This  is  effected  by  comparing  the  vibrations  of  the 
body  with  those  of  a  clock.  The  time  of  a  single 
vibration  or  of  any  small  arbitrary  number  of  vibra- 
tions cannot  be  observed  directlv,  because  this  would 
require  the  fraction  of  a  second  of  time,  as  shown  by 
the  clock,  to  be  estimated  either  by  the  eye  or  ear. 
The  vibrations  of  the  body  may  be  counted,  and  here 
there  is  no  fraction  to  be  estimated,  but  these  vibra- 
tions will  not  probably  fit  in  with  the  oscillations  of 
the  clock  pendulum,  and  the  differences  must  be  esti- 
mated. This  defect  is  overcome  by  "  the  method  of 
coincidences."  Supposing  the  time  of  vibration  of 
the  clock  to  be  a  little  less  than  that  of  the  body,  the 
pendulum  of  the  clock  will  gain  on  the  body,  and  at 
length  at  a  certain  vibration  the  two  will  for  an  in- 
stant coincide.  The  two  pendulums  will  now  be  seen 
to  separate,  and  after  a 'time  will  again  approach  each 
other,  when  the  same  phenomenon  will  take  place. 
If  the  two  pendulums  continue  to  vibrate  with  per- 


186  KEY  AND  SUPPLEMENT 

feet  uniformity,  the  number  of  oscillations  of  the  pen- 
dulum of  the  clock  in  this  interval  will  be  an  integer, 
and  the  number  of  oscillations  of  the  body  in  the 
same  interval  will  be  less  by  one  complete  oscillation 
than  that  of  the  pendulum  of  the  clock.  Hence  by 
a  simple  proportion  the  time  of  a  complete  oscillation 
may  be  found. 

The  coincidences  were  determined  in  the  following 
manner :  Certain  marks  made  on  the  two  pendulums 
were  observed  by  a  telescope  at  the  lowest  point  of 
their  arcs  of  vibration.  The  field  of  view  was  limited 
by  a  diaphragm  to  a  narrow  aperture  across  which  the 
marks  were  seen  to  pass.  At  each  succeeding  vibra- 
tion the  clock  pendulum  follows  the  other  more 
closely,  and  at  last  the  clock-mark  completely  covers 
the  other  during  their  passage  across  the  field  of  view 
of  the  telescope.  After  a  few  vibrations  it  appears 
again  preceding  the  other.  The  time  of  disappear- 
ance was  generally  considered  as  the  time  of  coinci- 
dence of  the  vibrations,  though  in  strictness  the  mean 
of  the  times  of  disappearance  and  reappearance  ought 
to  have  been  taken,  but  the  error  thus  produced  is 
very  small.  (Encyc.  Met.,  Figure  of  the  Earth.) 
In  the  experiments  made  in  Hartan  coal-pit  in  1854, 
the  Astronomer  Royal  used  Kater's  method  of  ob- 
serving the  pendulum.  (Phil.  Trans.,  1856.) 

The  value  of  T  thus  found  will  require  several  cor- 
rections. These  are  called  "  Reductions."  If  the 
centre  of  oscillation  does  not  describe  a  cycloid,  al- 
lowance must  be  made  for  the  alteration  of  time  de- 
pending on  the  arc  described.  This  is  called  "the 
reduction  to  infinitely  small  arcs."  If  the  point  of 


TO  ELEMENTARY  MECHANICS.  187 

support  be  not  absolutely  fixed,  another  correction  is 
required  (Phil.  Trans.,  1831).  The  effect  of  the 
buoyancy  and  the  resistance  of  the  air  must  also  be 
allowed  for.  This  is  the  "  reduction  to  a  vacuum." 
The  length  L  must  also  be  corrected  for  changes  of 
temperature. 

The  time  of  an  oscillation  thus  corrected  enables  us 
to  find  the  value  of  gravity  at  the  place  of  observa- 
tion. A  correction  is  now  required  to  reduce  this  re- 
sult to  what  it  would  have  been  at  the  level  of  the 
sea.  The  attraction  of  the  intervening  land  must  be 
allowed  for  by  Dr.  Young's  rule  (Phil.  Trans.,  1819). 
We  thus  obiain  the  force  of  gravity  at  the  level  of 
the  sea,  supposing  all  the  land  above  this  level  were 
cut  off  and  the  sea  constrained  to  keep  its  present 
level.  As  the  sea  would  tend  in  such  a  case  to  change 
its  level,  further  corrections  are  still  necessary  if  we 
wish  to  reduce  the  result  to  the  surface  of  that  sphe- 
roid which  most  nearly  represents  the  earth.  (See 
Camb.  Phil.  Trans.,  vol.  x.) 

There  is  another  use  to  which  the  experimental  de- 
termination of  the  length  of  a  simple  equivalent  pen- 
dulum may  be  applied.  It  has  been  adopted  as  a 
standard  of  length  on  account  of  being  invariable  and 
capable  at  any  time  of  recovery.  An  Act  of  Parlia- 
ment (5  Geo.  IV.)  defines  the  yard  to  contain  thirty- 
six  such  parts,  of  which  parts  there  are  39.1393  in  the 
length  of  the  pendulum  vibrating  seconds  of  mean 
time  in  the  latitude  of  London,  in  vacuo,  at  the  level 
of  the  sea,  at  temperature  62°  F.  The  commission- 
ers, however,  appointed  to  consider  the  mode  of  re- 
storing the  standards  of  weight  and  measure  which 


188  KEY  AND  SUPPLEMENT 

were  lost  by  fire  in  1834,  report  that  several  elements 
of  reduction  of  pendulum  experiments  are  yet  doubt- 
ful or  erroneous,  so  that  the  results  of  a  convertible 
pendulum  are  not  so  trustworthy  as  to  serve  for  sup- 
plying a  standard  for  length  ;  and  they  recommend 
a  material  standard,  the  distance,  namely,  between 
two  marks  on  a  certain  bar  of  metal  under  given  cir- 
cumstances, in  preference  to  any  standard  derived 
from  measuring  phenomena  in  nature.  (Report,  1841.) 

All  nations,  practically,  use  this  simple  mode  of 
determining  the  length  of  the  standard  of  measure, 
that  of  placing  two  marks  on  a  bar,  and  by  a  legal 
enactment  declaring  it  to  be  a  certain  length. 

For  length  of  seconds  pendulum  see  Mech.  Celeste^ 
T.  II.,  pp.  327,  343,  479. 

SOLUTIONS  OF  EXAMPLES. 

PACK  248. 

1.  From  the  equation  on  page  243  of  the  text  we 
have 


but  in  the  example  t  —  |  ;  hence 


471*  ~~  4  x  (3-1416)3 
0-827  feet. 
9-77  inches. 


TO  ELEMENTARY  MECHANICS.  189 

2.  For  this  example  we  have 


ff 


*»  ~  (3-1416)2 
=  13-036  feet. 

3.  First  find  the  time,  in  seconds,  of  one  vibration. 

In  one  day  there  are  24  X  60  X  60  =  86,400 
seconds  ;  hence  the  time  of  one  vibration  will  be 

I815S  =  0-99976  seconds. 

Let  x  be  the  required  length,  then  from  the 
equation  on  page  243  (f  being  the  time  for 
length  x)  we  have 

A  .   fV  .    .  ^  .    7 

(         *     V          •     •     M/     •     V} 

or 

1  :  (0-99976)2  : :  x  :  39-1 ; 
394 

•••*  =  (WJW6F=39-1181  +  inche8; 
hence  it  must  be  elongated 

39-1181  -  39-1  =  0-0181  +  inches. 

PAGE  249. 

4.  From  the  equation  on  page  246  of  the  text  we 

have 

,         45-5 

fl  =  T 

86354-5 

AK.  K 

20,923,161  feet, 


86354-5 
=  11024-3  feet. 


190  KEY  AND  SUPPLEMENT 

5.  From  the  equation 


5t  =  -, 

9 

we  have 

5t  =  15-7080  /A 
=  3-9168  seconds. 
6.  From  Eq.  (2),  page  246,  we  have 


=  A/32-0902  x  20,923,161 
=  25911-93  feet  per  second 
=  4-9  +  miles  per  second. 

7.  We  will  have  from  Eq.  (1),  page  246  of  the  text, 


=  3-1416  I/20-923-161 
32-0902 

=  42  in.  17  sec. 

8.  From  Equations  (1)  and  (4),  pages  247  and  248, 
we  have 


Fl  A/Ekg 
=       V- 


PEk 


TO  ELEMENTARY  MECHANICS. 
=  3000 


1000  x  28000000  x 
=  0-453  ft.  per  sec. 

9.  From  Eq.  (6)  we  have 


9 


J          1000  x  5 


416  r    32^  x  280(JOOOO  x  ^ 
=  0-0148  of  a  second. 

SOLUTIONS  OF  EXAMPLES. 
PAGE  258. 

1.  In  this  example  the  weight  of  the  fluid  is  ab- 
stracted, and  we  have  only  to  consider  the 
effect  of  the  pressure  of  the  piston.  The  area 
of  the  piston  is  ^nd?  =  0-7854  inches ;  hence 
the  pressure  p  per  square  inch  will  be 

p  =  20  -r-  0-7854  =  25-46  +  Ibs. 

The  area  of  the  bottom  of  the  box  =  2  x  3  x 
144  =  864  sq.  inches.  The  area  of  the  sides 
=  (2  x  1  x  2  +  2  x  1  x  3)  144  =  1440  sq. 
inches.  Hence  the  entire  area  of  the  bottom 
and  sides  will  be  2,304  sq.  inches ;  hence  the 
pressure  will  be 

P  =  2304  x  25-46 
=  58659-8  Ibs. 


192  KEY  AND  SUPPLEMENT 

2.  It  will  equal  the  weight  of  the  water  ;  hence 


=  1-736  Ibs. 

3.  The  area  of  the  base  will  be  3  1416  x  16.     The 

pressure  due  to  the  liquid  will  equal  the  weight 
of  a  cylinder  of  water  whose  base  is  8  inches  in 
diameter  and  height  10  inches,  or 

3-1416  x  16  x  10      C01 

-  —r=  -  x  62|  =  18-18  Ibs. 

1  I  -O 

The  pressure  upon  the  base  due  to  the  exter- 
nal pressure  of  100  Ibs.  will  be 


~  x  100  =  177-777  Ibs. ; 


hence  the  total  pressure  will  be 

P  =  18-208  +  177-777 
=  195-96  Ibs. 

4.  The  upward  pressure  equals  the  weight  of  water 
displaced — or  63  Ibs. ;  hence  the  pull  on  the 
string  will  be  the  difference  of  the  upward 
pressure  and  the  weight  of  the  block,  or 

63  -  35  =  28  Ibs. 
6.  We  have  from  Article  86,  p.  44  of  the  text, 


TO  ELEMENTARY  MECHANICS.  193 

FV&         28  x 


•~Jf~35  35 

And,  from  Eq.  (3),  p.  12  of  the  text, 


_  ./28  X  64^  x  50 

35 
=  50-73  ft.  per  sec. 

PAGE  259. 

G.  We  have,  for  the  volume  of  the  block  (|-)8  =  V- 
feet.  Hence  the  weight  =  V-  x  180  =  607-5 
Ibs.  The  upward  pressure  of  the  water  will 
be  ¥  x  62|  Ibs.  =  210-937  Ibs.,  which,  sub- 
tracted from  the  weight,  will  equal  the  tension, 
or  396-563  Ibs. 

7.  Let  a  equal  half  the  length  of  the  bar,  and  x  the 
distance  of  the  point  of  attachment  from  the 
middle  of  the  bar,  which  point  will  be  in  the 
iron  part.  Let  s  be  the  ratio  of  the  weight  of 
a  given  volume  of  wood  to  that  of  an  equal 
volume  of  water.  If  the  weight  of  the  water 
displaced  per  unit  of  length  of  the  bar  be 
called  unity,  then  will  the  weight  of  the 
wooden  part  of  the  bar  be  represented  by  sa, 
and  of  the  iron  part  by  8(ta.  The  upward 
pressure  of  the  water  will  be  2«,  since  it  equals 
the  volume  displaced.  The  resultant  upward 
stress  on  the  wooden  part  will  be  a  —  sa  = 
(1  —  s)a ;  and  its  moment  will  be  (1  —  s)a  x 
(a  +  ac).  The  resultant  downward  force  of  the 


194  KEY  AND  SUPPLEMENT 

bar  between  the  middle  of  the  beam  and  point 
of  attachment  will  be  (8*  —  l)a?,  and  the  mo- 
ment will  be  4(86-  —  l)^-2.  The  moment  of  the 
remaining  part  will  be  i(Ss  —  l)(a  —  a:)2. 
Hence  we  have  the  equation 

i(8*  -  l)(a  -  xf  =  fl(l  -  s)(a  +  x)-  $(Ss 


7*  ±  V-  111*-2  +68*-  6 

•'•  x  =  ~    -  Tc~~  —  a 

16s  —  2 

8.  If  one  end  is  depressed  3  inches,  the  other  will 
be  raised  the  same  amount,  hence  the  differ- 
ence of  level  will  be  6  inches,  and  the  column 
of  water  necessary  to  produce  this  difference 
of  level  will  be 

6  x  134-  =  81  inches. 

ANSWERS    TO    EXERCISES. 

1.  It  would  not  disperse,  but  would  remain  in  the 

same  form. 

2.  The  gas  would  till  the  hollow  space. 

3.  If  the  sides  of  the  pail  were  vertical,  it  probably 

would  ;  but  even  in  this  case,  if  it  were  poured 
in  and  run  down  the  side,  it  would  require 
a  short  time  to  overcome  the  adhesion  on  the 
side  and  permit  the  entire  weight  to  be  exerted 
on  the  bottom  of  the  vessel. 

4.  As  much  less  as  the  weight  of  an  equal  volume 

£        '  ' 

of  air. 

5.  Of  the  same  density  of  air  —  or  the  weight  must 

be  the  same  as  that  of  an  equal  volume  of  air. 


TO  ELEMENTARY  MECHANICS.  195 

SOLUTIONS   OF    EXAMPLES. 


PAGE  270. 

(S2  —  *)•<?! 

1.  u\  —     -  — w. 

s  _ 


+  ws%  —  ws 
substituting  the  value  of  s  =  1-3077,  we  have 

12  x  1-3077  x  11  j 

~  34  x  11  +  (12  -  34)1-3077  " 

w  32  32 

*>  Q   .    .    A  4 

***         °   ~~  —  on  c\f  —      rr      —   -^T" 

w  —  wl       32  —  25       7 

r  w        I       7/' 

3.     s  =  -™ -.  In  this  case  wl  =  0 ;  .'.  solving  for 


^  -8  x  60      48       0/m 

=          =:  :=: 


4.  A  stone  5  ft.  on  each  edge  =  125  cu.  ft.  and  will 

displace  125  x  62-5  =  7812-5  fts.  water,  x  2-3 
=  17968-75  Bbs.  =  weight  of  stone. 

5.  «1==f 

w  —  u'2      40-32      8 

*•       «  =  -  —  T7J  -  TF—  F  =  !*"• 

w  —wt      40  —  35      5 
7  w  +  V&        35  +  18      53  _ 

c  +  «!        '5+2      "  7  " 
-(*-*>!,.   _  (10-5-  14)19-3  x  10  _ 
(*2-*i)V  (10-5-19-3)14 

-675-5 

=  5'483' 


196  KEY  AND  SUPPLEMENT 

_  (sl  -  8)%       _  (19-3  -  14)10-5  x  10  _ 
**  ~  (*i  -  *8)*  '  (19-3  -  10-5)14 

556-5 


o    1  -  i  _  P*  +  PI*I  _  I  _  27  x  1  +  39  -4915  x  1-8  ±85 
'  n  =  (o  +  Vfa  ~  (27  +  39-4915)1-6321 


_  ft,  +  0-0013  (ft,  -  c)  _  14  +  0-0013(10  -7) 
flj  +  *2  -  c  14  +  10-7 

14-0039 


ANSWERS  TO  EXERCISES. 

PAGE  271. 

1.  It  will  not  ;  for  water  being  more  compressible 

than  the  solid,  will  be  relatively  more  dense 
in  the  air  than  in  a  vacuum,  and  hence  when 
in  air  will  force  the  body  upward  more  than 
when  in  a  vacuum. 

2.  Because  the  smoke  is  lighter  than  the  surround- 

ing air  ;  but  if  it  be  heavier  it  will  fall  in  the 
air. 

3.  See  answer  to  Exercise  1. 

4.  This  assumes  that  the  weight  of   the  bag  —  or 

some  other  cause  —  causes  the  bag  to  sink,  and 
if  it  sinks  at  all  it  will  go  to  the  bottom  of  the 
vessel.  If  now  a  pressure  be  exerted  upon  the 
surface  of  the  liquid,  it  will  cause  the  bag  (or 


TO  ELEMENTARY  MECHANICS.  197 

gas)  to  condense  more  than  the  liquid,  and 
lience  it  will  not  rise.  Toys  have  been  made 
involving  this  principle. 

5.  Water  is  more   compressible  than  iron  for  the 

same  pressure,  hence  it  seems  possible,  theoret- 
ically, for  water  to  be  subjected  to  such  a  pres- 
sure as  to  be  as  dense  as  iron  at  the  same  pres- 
sure. 

6.  If  both  are  incompressible,  their  relative  densi- 

ties will  be  unchanged  by  pressure,  and  hence 
the  heavier  body  will  sink  indefinitely.  If  the 
body  be  compressible,  it  will  become  relatively 
more  dense,  and  hence  there  will  be  no  limit. 

7.  If  the  brine  be  sufficiently  "  strong  " — according 

to  popular  language — it  will  float  the  egg.  In 
order  that  it  may  float  between  the  top  and 
bottom,  the  brine  must  be  more  dense  near  the 
bottom  than  at- the  top. 

8.  It   will.     It   is   related  of   Benjamin  Franklin, 

that  he  asked  a  company  of  savants  why  a  pail 
of  water  containing  a  fish  would  weigh  no  more 
than  without  the  fish.  Several  reasons  were 
given,  and  finally  he  was  appealed  to  for  the 
reason.  He  thus  replied:  "Are  you  sure  it 
will  weigh  no  more?"  They  had  been  trying 
to  explain  a  false  assumption. 

SOLUTIONS    OF   EXAMPLES. 

PAGE  2T6. 

f         ? 

1.  The  equation  on  page  273  gives  us  -  = 

*/ 

tan    > 


198  KEY  AND  SUPPLEMENT 

tan  <p  =  fg  =0-3  ;  .:<?=  16°  41'  57". 

2.  f—  g  tan  <p.  The  section  of  the  liquid  will  be 
a  triangle  with  a  base  of  3  ft.  and  a  height  of 
2  ft.,  hence  tan  cp  =  f  ; 

/./=  f  x  32£  =  21$  feet  per  second. 

&.f=g  tan  q>.  In  this  case  the  slope  of  the  free 
surface  will  be  45°,  and  tan  45°  =  1  ;  .\f=y. 

4.  Let  A  be  the  edge  of  the  vessel.  From  A  con- 
ceive a  horizontal  line  drawn  to  XE,  and  mark 
the  foot  with  the  letter  Z.  The  volume  gen- 
erated by  the  revolution  of  the  semi-parabola 
is  one-half  the  product  of  the  base  and  altitude 
(See  Mensuration^  or  works  on  the  Integral 
Calculus).  And  as  the  volume  of  this  parabo- 
loid is  the  unoccupied  portion  of  the  cylinder, 
the  altitude  ZE  will  be  2  x  3  =  6  inches.  AZ 
is  12  inches.  From  a  property  of  the  parabola, 
we  have 

EZ  :  AZ  :  :  AZ  :  the  parameter, 
(orx:y::y:  2/>)  ; 

122 

.-.  the  parameter  =  -^  =  24  inches. 

This  is  known  to  be  twice  the  subnormal  DC] 
hence  DC  =  12  inches,  which,  as  shown  on 
p.  275  of  the  text,  is  g  -r-  or2,  therefore 


. 

12 


TO  ELEMENTARY  MECHANICS.  199 


Let  n  be  the  number  of  turns  per  minute 
sought.  Then  2?r  —  =  —  —  will  be  the  angular 
velocity  per  second,  and  we  have 


30      "  12  ' 
.•.n  =  11-77  turns. 

5.  The  angular  velocity  will  be  f  #2;r  =  ?r,  which  is 
the  value  of   GO  in  the  value  of  DO,  p.  275  of 

the  text  ;  hence  DC  —  ~s-.     This    is   one-half 

71 

the   parameter    in    the    equation   y*  = 
hence  the  equation  becomes  y*  —  2  -^  x. 

SOLUTIONS  OF  EXAMPLES. 

PAGE  282. 

1.  To  fulfill  this  condition  we  must  have 


-  A2) 
A2  =  4-5 
h  =  2-121  feet. 

2.  The  area  of  each  triangle  will  be  %  x  1-4  x  2-6 
=  1-82.  The  centre  of  gravity  of  the  triangle 
whose  base  will  be  in  the  surface  will  be 
|  x  2-6  x  sin  56°  35'  below  the  free  surface; 
hence  the  pressure  on  it  will  be 


x  1-82  x  i  x  2-6  x  sin  56°  35'  =  82-28  Ibs.  ; 
and  of  the  other  triangle, 


200  KEY  AND  SUPPLEMENT 

62|  x  1-82  x  f  x  2-6  x  sin  56°  35'  =  164-57  Ibs. 
3.  Pressure  on   concave  surface  =  £#&/i2,  but  b  = 


.-.p  =  A.  x  62-5  x  2  x  34416  x  9  =  1767-15  Ibs. 

Weight  of  liquid  =  TV*  x  A  x  6  =  3-1416  x  3 

x  62-5  =589-05  fts. 

Pressure  on  base  =  weight  of  liquid  =  589-05 
fts. 

4.  The  weight  of  the  liquid  =  ^TT^  =  f  x  62-5  x 

3-1416  x  125  =  32725-0  fts. 
Normal  pressure  =  4#7r/-3  =  4  x  62'5  x  3-1416 
x  125  =  98175  fts. 

5.  Pressure  on  flood-gate  =  i<W>(/i23  —  ^'i2)  =  i  x 

62-5  x  2  x  (132  -  TO2)  =  62-5  x  69  =  4312-5  fts. 


6.  Pressure  on  opposite  side  —  £#5(7i'2  —  A"2)  =  ^  x 
2  x  62-5(72  -  42)  =  62-5  x  33  =  2062-5, 

and 
4312-5  -  2062-5  =  2250  fts. 

SOLUTIONS    OF    EXAMPLES. 

PAGE  287. 

1.  By  Art.  372,  the  centre  of  pressures  of  rectangle 

is  at  a  distance  from  the  top  equal  to  f  the 
height,  .-.  |  of  3  =  2  feet. 

2.  By  Art.  373,  we  have  for  the  required  depth, 

35      1.5  /-  +  ION      35       5     19      1365       95 
* 


6       36    V  +   5     ~~  6       J8    13  ~  234       234 


TO  ELEMENTARY  MECHANICS.  201 


=  5-427  ft;  or  -427ft.  =  519  inches 
below  the  top  of  the  flood-gate. 

3.  To  resist  overturning  we  have 

Stfh  =  1f1/V  substituting  the  values  given, 
and  we  have 

„  _  125  x  512    ._  64000  _  . 

6  x  180  x  8       8640"  " 


.-.  &  =  V7-4074  =  2-72  ft.  =  2  feet  8|  inches. 


4.  In  this  case  we  have,  p.  279  of  the  text, 

fibs.  for  the  pressure  of  the  liquid.  Its  mo- 
ment in  reference  to  the  edge  of  the  wall  will 
be  £  x  31'|M3,  and  the  wall  must  be  capable 
of  resisting  twice  this  amount.  The  moment 
of  the  wall  will  be 

i  x  4  x  8  x  120  x  |-  of  4  =  5120  Ibs.  ; 
.-.  |  x  31$  JA8  =  5120  ; 
/.  h  =  6-2  feet. 

5.  The  pressures  are  proportional  to  the  areas,  and 

the  areas  are  as  (15)2  -j-  (1'5)2  =  100  to  1  ; 
.-.  total  pressure  =  500  x  100  =  50,000  Bbs. 

SOLUTIONS  OF  EXAMPLES. 
PAGE  310. 

1    t=      2^_= 


0-62  x  7r(-4\-)V64i  x  3 
9* 


202  KEY  AND  SUPPLEMENT 

2x1x3x2304 

=  -  —  =  401-2  sec.  =  6m.  41-2  sec. 

0-62V193 

2.  Q  =  \mbt  Vfyh3  =  |  x  06-2  x  2  (45  x  60) 


x  (f)3 
=  13618-8  cubic  feet. 

3.  The   equation   on   p.    309    of  the  text  is  x  = 
-5V4-     But  h  =  24  inches,  and  r  =  3  inches, 

rp*  " 

hence  we  have  a;  =  —  y*  =  —  y4. 

To  find  the  area  of  the  orifice,  we  have  on 
p.  309  of  the  text, 

nv*c  h         2          1 

z 


=  I-  ft.,  A  =  2  ft.  ; 
3-1416  x        x 


, sq.  f  t. 

0-62 V64£  x  2 

=  0-01341  sq.  in. 
4.  From  the  equation  on  p.  303  of  the  text  we  have 

25  -0-025187-^  =  0-0006769 
\k) 


or 

25  -  32-64235^=  7895-36(^  +  0.0039360  ; 
.  &   31-076  Q   J>5_, 
7928  V   7928' 
/.  Q  =  0-0542  en.  ft.  per  sec. 
=  195-1  +  cu.  ft.  per  hour. 


TO  ELEMENTARY  MECHANICS.  203 

ANSWERS  TO  EXERCISES. 
PAGE  310. 

1.  The  time  will  be  the  same  for  each. 

2.  It  will  not.     The  time  will  be  less,  for  the  head 

producing  the  velocity  will  be  equivalent 
to  what  it  would  be  if  for  the  weight  of  the 
water  an  equal  weight  of  mercury  be  substi- 
tuted. 

3.  It  will  not.     The  flow  of  the  water  in  this  case 

will  exceed  that  of  the  mercury  in  the  preced- 
ing. 

4.  It  will.     It  may  be  observed  that  when  the  pres- 

sure producing  the  flow  of  a  liquid  is  the  weight 
of  the  same  liquid,  the  head  equals  the  height 
of  the  liquid  above  the  orifice. 

5.  It  will  not,  but  the  depth  of  submergence  will 

gradually  increase.  The  block  receives  an  ini- 
tial velocity  downward  which  is  being  gradu- 
ally lessened  as  the  surface  of  the  liquid  de- 
scends. 

6.  It  will  be  lowest  near  the  orifice. 

7.  It  will  be  greater ;  for  the  acceleration  upward 

of  the  vessel  will  have  the  same  effect  as  an 
increased  pressure  on  the  surface. 

8.  It  will  be  greater ;  for  the  head  over  the  orifice 

will  be  greater. 

9.  It  is  not ;  a  part  of  the  pressure  is  engaged  in 

producing  motion  of  the  mass. 


204  KEY  AND  SUPPLEMENT 

PAGES  326-329. — The  expression  for  the  pressure  (or 
rather  the  tension)  of  the  air  at  any  height,  a?,  above 
the  earth,  p.  326,  reduces  to 

_x        <n 

p  =  p0e    »=£f-j 

eff 

where  p0  =  15  Ibs.,  e  =  2-71828  +,  and  H=  26214 
feet,  according  to  Rankine ; 

15 


(2-71828)26214 

If  x  =  20,000,000  feet,  about  the  radius  of  the  earth, 
we  have 

15  15 


(2-71828)*-       S 
nearly. 

But  this  expression  gives  too  rapid  a  diminution 
of  the  tension,  since  the  effect  of  gravity  is  discarded. 
Newton,  in  the  Principia,  gives  the  following: 

PROPOSITION  xxii.,  B.  II. — Let  the  density  of  any 
fluid  be  proportional  to  the  compression,  and  its  parts 
attracted  downwards  by  a  gravitation  reciprocally 
proportional  to  the  squares  of  the  distance  from  the 
centre.  I  say,  that  if  the  distances  be  taken  in  har- 
monic progression ,  the  densities  of  the  fluid  at  those 
distances  will  be  in  geometrical  progression.  In  ac- 
cordance with  this  proposition  the  following  table 
was  computed  in  The  System  of  the  World  by  Sir 
Isaac  Newton : 


TO  ELEMENTARY  MECHANICS. 


205 


COMPRESSION     OF     THE    AIR. 

HEIGHT  IN  MILFS. 

INITIAL  PRESSURE  EQUAL 
A  COLUMN   OF   WATER   33 

EXPANSION   OP  THE  AIR,  THE 
INITIAL      VOLUME       BEING 

FEET  HIGH. 

UNITY. 

0 

33 

1 

5 

17.1815 

1.8486 

10 

9.6717 

3.4151 

20 

2.852 

11.571 

40 

0.2525 

136.83 

400 

0.(10I7)1224 

26956  x  10'» 

4.000 

O.(10ld6)465 

73907  x  1010* 

40,000 

O.(10"")1628 

20263  x  10"" 

400,000 

O.(10210)7895 

41798  x  10207 

4,000,000 

O.(10212)9878 

33414  x  102U" 

Infinite. 

0.(102I2)6041 

54622  x  1020» 

where  (1017)1224,  implies  that  there  are  17  cyphers 
before  1224:  in  the  decimal,  thus  0  000000000000000- 
001224-,  and  similarly  for  the  others.  The  following 
inference  is  drawn :  "  But  from  this  table  it  ap- 
pears that  the  air,  in  proceeding  upwards,  is  rare- 
fied in  such  a  manner,  that  a  sphere  of  that  air 
which  is  nearest  to  the  earth,  of  but  one  inch  in 
diameter,  if  dilated  with  that  rarefaction  which  it 
would  have  at  the  height  of  one  semi-diameter  of  the 
earth,  would  fill  all  the  planetary  regions  as  far  as  the 
sphere  of  Saturn,  and  a  great  way  beyond  ;  and  at 
the  height  of  ten  semi-diameters  of  the  earth  would 
till  up  more  space  than  is  contained  in  the  whole 
heavens  on  this  side  the  fixed  stars,  according  to  the 
preceding  computation  of  their  distance.  And  though, 
by  reason  of  the  far  greater  thickness  of  the  atmos- 
pheres of  comets,  and  the  great  quantity  of  the  circum- 
solar centripetal  force,  it  may  happen  that  the  air  in 
celestial  spaces,  and  in  the  tails  of  comets,  is  not  so 


206  KEY  AND  SUPPLEMENT 

vastly  rarefied,  yet  from  this  computation  it  is  plain 
that  a  very  small  quantity  of  air  and  vapor  is  abund- 
antly sufficient  to  produce  all  the  appearances  of  the 
tails  of  comets,  for  that,  they  are  indeed  of  a  very 
notable  rarity  appears  from  the  shining  of  the  stars 
through  them."  Similar  remarks  are  made  in  the 
Principia  under  PROP.  xli. 

SOLUTIONS  OF  EXAMPLES. 
PAGE  331. 

1.  In  the  first  formula  on  p.  317  of  the  text,  mak- 

ing Fj  =  2  F0  and  a  =  0-002039,  we  find 

*  ry~  OTKS5S9  =  49°-  + 

PAGE  332. 

2.  At  the  surface  of  the  earth  the  pressure  of  the 

air  will  balance  a  column  of  water  34  feet 
high.  The  pressures  will  be  inversely  as  the 
amount  of  compression  ;  hence 

I  :  30  :  :  34  :  x  -  1020  feet  of  water. 

But  the  first  34  is  due  to  atmospheric  pressure  ; 
hence  the  depth  will  be  1020  -  34  =  986  feet. 

3.  From  the  equation  in  Prob.  3,  p.  323  of  the  text 

we  have 


=  0-322  of  an  inch. 

4.  Assuming  that  a  cubic  foot  of  air  weighs  0-08072 
of  a  pound,  10,000  cubic  feet  will  weigh  807-2 


TO  ELEMENTARY  MECHANICS.  207 

Ibs.,  and  the  weight  of  the  gas  will  be  807-2  x 
0-069  =  55-696  Ibs.  This  subtracted  from  the 
weight  of  an  equal  volume  of  air  will  give  the 
lifting  capacity,  or  807-2  -  55-696  =  741-504 
Ibs. 

5.  From  the  last  equation  of  Article  419,  p.  317  of 
the  text,  we  have  (t0  being  32°) 


A      1  +  0-002039(400  -32) 
~  5-5  X  1  +  0-002039(32  -  32) 
=  23-868  Ibs. 


Why  will  a  wheel  with  nearly  all  the  matter  con- 
centrated in  the  axle  roll  down  a  plane  in  less  time 
than  if  it  be  nearly  all  concentrated  in  the  rim  ? 

Because,  in  descending  the  plane  the  entire  work 
is  done  by  gravity,  and  the  measure  of  that  work  is 
the  weight  into  the  vertical  descent  of  the  centre  of 
the  wheel ;  and  this  work  is  changed  into  kinetic  en- 
ergy in  the  wheel.  When  the  matter  is  concentrated 
in  the  axle,  the  energy  will  be  ^mv*,  where  m  is  the 
mass,  and  v  the  final  velocity  of  the  centre ;  but  if  it 
be  all  concentrated  in  the  rim  its  energy  will  be 
$mvi  +  |7ft(rc»)2,  where  in  is  the  same  mass  as  before, 
Vi  the  final  velocity  of  the  centre,  GO  the  final  angular 
velocity  of  the  rim,  and  r  the  radius  of  the  rim. 
Since  the  total  kinetic  energy  must  be  the  same  in 
both  cases,  it  is  evident  that  v  must  exceed  v^  and 
hence  the  time  of  descent  in  the  latter  case  will  ex- 


208  KEY  TO  ELEMENTARY  MECHANICS. 

ceed  that  in  the  former.  It  is  shown  in  Analytical 
Mechanics,  p.  215,  that  if  a  given  mass  be  entirely  con- 
centrated in  the  axle,  or  uniformly  distributed  as  a  disc, 
or  entirely  concentrated  in  the  rim,  all  having  the 
same  radius,  the  times  will  be  as  V2  :  \/3  :  A/4. 


THE   END. 


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